Prob Set 1 PMAT 507 PROBLEM MATH

Darius L. Fajardo

Problem Set No. 1 PMAT 507 |1

Problem 1

Explain Newton’s law of cooling. Upon what factors does the rate of cooling ( warming) of an object depend?

ANSWER: Newton's Law of Cooling states that the rate of change of the temperature of an object is proportional to the difference between its own temperature and the ambient temperature (i.e. the temperature of its surroundings). Below is the equation form of the law and its corresponding solution Newton’s Law of Cooling  t

= – k(T k(T –  – Ta); T(0) = To

Solution  – kt

T(t) = T a + (To – Ta)e Where:

dT/dt = rate of change in temperature, C°/s Ta = ambient temperature, °C To = initial temperature of the body, °C t = later time, s  – 1 k = constant value for rate of cooling, s

Darius L. Fajardo

Problem Set No. 1 PMAT 507 |2

Problem 2

On a linear temperature scale  Z , water boils at  – 53.5°Z and freezes at  – 170°Z. What is a temperature of 350 K on the Z scale? GIVEN: Boiling point of water in  Z scale, – 53.5°Z Freezing point of water in  Z scale, – 170°Z Boiling point of water in Celsius scale, 100°C Freezing point of water in  Z scale, 0°C Temperature, T = 350 K REQUIRED: Equivalent of 350 K to  Z scale, t SOLUTION: By linear interpolation between  Z scale and Celsius scale - . - (-) t - (-) . t  





 -  t - 

 t

------------------ equation 1

T = t°C + 273.15 t°C = T – 273.15 ------------------ equation 2 Substitute equation 2 to equation 1 . t  



  - .

  t   -  .  - .  -  t   .  - .

t°Z = – 80.5°Z ANSWER: t°Z = – 80.5°Z

Darius L. Fajardo

Problem Set No. 1 PMAT 507 |3

Problem 3

Increasing the temperature of a copper coin by 100 C° makes its diameter increase by 0.18%. i) What is the percent increase in the coin’s a) area of a face? b) thickness? c) volume? d) mass? ii)Calculate the coefficient of linear expansion of the coin. GIVEN: hange in temperature, Δt =   Increase in diameter, ΔD/D o = 0.18% REQUIRED: i) a) increase in face area, ΔA/A o i) b) increase in thickness, ΔL/L o i) c) increase in volume, ΔV/V o i) b) increase in mass, Δm/m o

ii) coefficient of linear expansion of the coin, α

SOLUTION: i) a) Solving for ΔA/A o Since the coefficient of area expansion is twice the linear expansion, therefore ΔA/Ao = 2 ΔD/Do ΔA/Ao = 2 (0.18%) ΔA/Ao = 0.36 % i) b) Solving for ΔL/L o Since thickness is also a linear dimension, therefore ΔL/Lo = ΔD/Do ΔL/Lo = 0.18% i) c) Solving for ΔV/V o Since the coefficient of volume expansion is thrice the linear expansion, therefore ΔV/Vo = 3 ΔD/Do ΔV/Vo = 3 (0.18%) ΔV/Vo = 0.54 % i) b) Solving for Δm/m o Assuming no relativistic effect, mass is conserved, therefore Δm/mo = 0 ii) Solving for α ΔL = L oαΔt α=

α=

ΔL Lo Δt .  

 – 5

α = 1.8 x 10 /C° ANSWER:

i) a)ΔA/Ao = 0.36 % i) b) ΔL/Lo = 0.18% i) c) ΔV/V o = 0.54 % i) ) Δm/mo = 0  – 5 ii) α = 1.8 x 10 /C°

Darius L. Fajardo

Problem Set No. 1 PMAT 507 |4

Problem 4

A bar of length 3.77 m has a crack at its center. When its temperature rises by 32 C°, it buckles upward (see figure). The bar has linear expansivity coefficient of 2.50 x 10

 – 5

/C°. Find the rise  x of its center.

GIVEN: Initial length of bar, L o = 3.77 m hange in temperature, Δt =    – 5 Coefficient of linear expansion, α = . x  /C° REQUIRED: Rise x  SOLUTION: From the figure, solve for x x=

√ c - (.L ) 

o

2

------------------ equation 1

Solving for c 2c = Lo(  αΔ) c = ½ L o(  αΔ)  – 5 c = ½ (3.77 m)[1 + (2.50 x 10 /C°)(32C°)] c = 1.886508 m From equation 1

√ 

x =  -    2

x = 0.075 m ANSWER: x = 0.075 m

Darius L. Fajardo

Problem Set No. 1 PMAT 507 |5

Problem 5

A small electric immersion heater, rated at 200 W, is u sed to heat 100 g of water for instant coffee. How long will it take to boil the water which is at a temperature of 23.0°C? Ignore all heat losses. GIVEN: Electric power of heater, P = H = 200 W (100% efficient) Mass of water, m = 100 g; m = 0.100 k g Initial temperature of water, t o = 23.0°C Boiling point of water, t = 100°C REQUIRED: Time to boil the water, T SOLUTION: H=

  

--------------------------- equation 1

Solving for Q  Q = mC(t – to) ------------- equation 2 Substitute equation 2 to equation 1 H=

m(t - to ) 

Solving for T T=

T=

m(t - to )  (.)()( - )

T = 161 s ANSWER: T = 161 s



s

Darius L. Fajardo

Problem Set No. 1 PMAT 507 |6

Problem 6

A 150-g copper bowl contains 220 g of water at 20.0°C. A very hot copper cylinder, of mass 300 g, is dropped into the water, and this caused 5.00 g of steam at 100°C to form. Neglecting all heat energy transfer to the environment, determine the a) heat energy (calories) transferred to the water; b) heat energy (calories) absorbed by the bowl; c) initial temperature of the copper cylinder. GIVEN: Mass of copper bowl, m A = 150 g; m A = 0.150 kg Mass of water, m B = 220 g; mB = 0.220 kg Initial temperature of water and copper bowl, t oA = 20.0°C Mass of copper cylinder, m C = 300 g; mC = 0.300 kg Mass of steam, m D = 5.00 g; mD = 0.005 kg

Equilibrium temperature, t = 100°C Specific heat of water, C B = 4186 J/kg.C° Specific heat of copper, C A = 390 J/kg.C° Heat of vaporization of H 2O, LV = 2256 KJ/kg

REQUIRED: a) Heat energy transferred to the water, Q B b) Heat energy (calories) absorbed by the bowl, Q A c) Initial temperature of the copper cylinder, t oC SOLUTION: a) Solving for Q B Q B = mBCB(t – toA) + mDLV Q B = (0.22)(4186)(100 – 20) + 0.005(2256000) J 4 Q B = 8.50 x 10 J (1 cal/4.186 J) 4 Q B = 2.03 x 10 cal b) Solving for Q A Q A = mACA(t – toA) Q A = (0.15)(390)(100 – 20) J 3 Q A = 4.68 x 10 J(1 cal/4.186 J) 3 Q A = 1.12 x 10 cal c) Solving for t oC Q A + Q B + Q C = 0 --------------------------- equation 1 Solving for Q C Q C = mCCA(t – toC) ------------------------- equation 2 Substitute equation 2 to equation 1 Q A + Q B + mCCA(t – toC) = 0 toC = t –

(

      ) A

toC = 100 –

(-  - .)

t = 866°C ANSWER: 4 a) Q B = 2.03 x 10 cal 3 b) Q A = 4.68 x 10 cal c) t = 866°C



(m A )

(. x )

Darius L. Fajardo

Problem Set No. 1 PMAT 507 |7

Problem 7

Ethyl alcohol has these thermal properties: boiling point, 78.0°C; freezing point,  – 114°C; latent heat of  vaporization, 879 kJ/kg; latent heat of fusion, 109 kJ/kg, and specific heat, 2.43 kJ/kg. How much thermal energy is required to solidify (at  – 114°C) 0.525 kg of gaseous ethyl alcohol at a temperature of 78.0°C? GIVEN: Boiling point, t o = 78.0°C Freezing point,t = – 114°C Latent heat of vaporization, L V = 879 kJ/kg Latent heat of fusion, L f  = 109 kJ/kg Specific heat, C = 2.43 kJ/kg Mass of ethyl alcohol, m = 0.525 kg REQUIRED: Heat to solidify ethyl alcohol, Q  SOLUTION: Q = Q V + Q S + Q f  ------------------ equation 1 Solving for Q V Q V = – mLV --------------------------- equation 2 Solving for Q S Q S = mC(t – to) --------------------- equation 3 Solving for Q f  Q V = – mLf --------------------------- equation 4 Substitute equation 2, equation 3, equation 4 to equation 1 Q = – mLV + mC(t – to) + – mLf  Q = m[ – LV + C(t – to) + – Lf ] Q = 0.525[ – 879 + 2.43( – 114 – 78) + – 109] Q = – 764 kJ

ANSWER: Q = – 764 kJ

Darius L. Fajardo

Problem Set No. 1 PMAT 507 |8

Problem 8

In a solar water heater, solar radiation is absorbed by water that circulates in tubes in a roof top collector. Warm water is then pumped onto a holding tank. If the device is 20.0% efficient, what collector area is necessary to raise the temperature of 200 L of water in the tank from 20.0°C to 40.0°C in 1.00 hour when the intensity of incident 2 sunlight is 700 W/m ? GIVEN: Efficiency of heater, e = 20.0%; e = 0.20 3 Volume of water, V = 200 L; V = 0.200 m Initial temperature of water, t o = 20.0°C Final temperature of water, t = 40.0°C Time, T = 1.00 hr; T = 3600 s 2 2 H/A = 700 W/m ; H = (700 W/m ) A REQUIRED: Collector area, A SOLUTION: Q water = eQ heater ------------------ equation 1 Solving for Q water Q water = mC(t – to) -------------- equation 2 Solving for m m = ρV ------------------------------ equation 3 Substitute equation 3 to equation 2 Q water = ρV (t – to) ------------ equation 4 Solving for Q heater Q heater = Ht ------------------------- equation 5 Substitute equation 4, and equation 5 to equation 1 ρV (t – to) = e Ht 2 ρV (t – to) = e (700 W/m ) A t ρV(t to ) A=



(et)

A=

()(.)()(

( x . x )

A = 33.2 m ANSWER: 2 A = 33.2 m

 )

2

Darius L. Fajardo

Problem Set No. 1 PMAT 507 |9

Problem 9

A 20.0-g copper ring has a diameter of 2.54000 cm at 0.000°C. An aluminium sphere has a diameter of 2.54508 cm at 100.0°C. The sphere is place on top of the ring, and the two are allowed to come into thermal equilibrium, without any heat lost to the surroundings. At thermal equilibrium, the sphere just passes through the ring. What is the mass of the sphere? GIVEN: Mass of copper ring, m A = 20.0 g Specific heat of copper, C A = 390 J/kg.C° Initial diameter of ring, D OA = 2.54000 cm Specific heat of aluminum, C B = 910 J/kg.C° Initial temperature of copper ring, t OA = 0°C Initial diameter of aluminium sphere, D OB = 2.54508 cm Initial temperature of aluminium sphere, t OB = 100°C  – 5 oefficient of thermal expansion of copper, α A = 1.7 x 10 /C°  – 5 Coefficient of thermal expansion of aluminum , αB = 2.4 x 10 /C° REQUIRED: Mass of sphere, m B SOLUTION: Solving for t, DA = DB DoA*  α A(t – tOA)] = DoB*  α B(t – tOB)] t = [D oB(1 – αB tOB) + DoA(αA tOA – 1)] /(DoA αA – DoB αB)  – 5

 – 5

 – 5

 – 5

t = {2.54508[1 – (2.4 x 10 )(100)] + 2.5[(1.7 x 10 ) (0) – 1)] /[(2.54)( 1.7 x 10 ) – (2.54508)( 2.4 x 10

t = (2.538971808 – 2.54)/ -0.00001790192 t = 57.435°C Solving for mB Q A + Q B = 0 ------------------ equation 1 Solving for Q A Q A = mACA(t – toA) --------- equation 2 Solving for Q B Q B = mBCB(t – toB) --------- equation 3 Substitute equation 2, and equation 3 to equation 1 mACA(t – toA) + mBCB(t – toB) = 0 mB = – mACA(t – toA)/ CB(t – toB) mB = – (20)(390)( 57.435 – 0)/ (910)( 57.435 – 100) mB = 11.6 g

ANSWER: mB = 11.6 g

)]

Darius L. Fajardo

P r o b l e m S e t N o . 1 P M A T 5 0 7 | 10

Problem

Two identical rectangular rods of metal are welded end to end as shown in the upper figure and a steady-state heat transfer of 10 J occurred in 2.00 minutes. How long would the process last if the rods are welded together as shown in the lower figure? GIVEN: Heat conducted, Q = 10 J Time, TS = 2.00 minutes; T S = 120 s (for series) Cold side temperature, t C = 0°C Hot side temperature, t H = 100°C Thickness of rods, L REQUIRED: Time of conduction when the rods are parallel, T P SOLUTION: Analyzing series connection Solving for heat current of conduction f or series, H S HS =

  S

------------------ equation 1



A(  )

HS =

L k



kA(  )

HS =

L

------- equation 2

Equate equation 1 and equation 2   S

=



kA(  ) L

 



kA(  )

S

=

L

--------- equation 3

Analyzing parallel connection Solving for heat current of conduction f or series, H S HP =

HP =

  

------------------ equation 4



kA(  ) L

----- equation 5

Equate equation 4 and equation 5   

=



kA(  ) L

 



kA(  )

=

 L

------ equation 6

Darius L. Fajardo

P r o b l e m S e t N o . 1 P M A T 5 0 7 | 11

Equate equation 3 and equation 6  L

=

TP

=

TP

=

S L

S   s 

TP = 30.0 s ANSWER: TP = 30.0 s

Darius L. Fajardo

P r o b l e m S e t N o . 1 P M A T 5 0 7 | 12

Problem

A wall consists of four layers of different materials (see figure), of thermal conductivities as given: k 1 = 0.060 W/mK; k3 = 0.040 W/mK; k 4 = 0.120 W/mK; L1 = 1.50 cm; L 3 = 2.80 cm; L4 = 3.50 cm. What is the interface temperature indicated? GIVEN: k1 = 0.060 W/mK k3 = 0.040 W/mK k4 = 0.120 W/mK TH = 50.0°C TC = - 10.0°C T12 = 25.0°C

L1 = 1.50 cm L3 = 2.80 cm L4 = 3.50 cm

REQUIRED: Temperature at the interface of wall 3 and 4, T SOLUTION: Solving for H/A using wall 4  A

=



k (  ) L

------------------ equation 1

Solving for H/A using wall 1  A

=



k (  ) L

--------------- equation 2

Equate equation 1 and equation 2



k (  ) L

T=

T=

=

L



k L (  ) k L

+ TC

(.)(.)( - ) (.)(.)

T = 19.2°C

ANSWER: T = 19.2°C



k (  )

+ ( – 10)

Darius L. Fajardo

P r o b l e m S e t N o . 1 P M A T 5 0 7 | 13

Problem

A tank of water has been outdoors in a very cold weather and a slab of ice 5.00 cm thick has formed on its surface (see figure below). The air above the ice is at  – 10.0°C. Calculate the rate of formation of ice (in cm/hr) on the ice slab. The thermal conductivity and density of ice are 0.00400 cal/cm.c.C° and 0.920 g/cc, respectively. No heat energy is transferred through? GIVEN: Cold side temperature, T C = – 10.0°C Hot side temperature, T C = 0°C Latent heat of fusion of ice, L F = 79.6 cal/g Thermal conductivity ice, k = 0.00400 cal/cm.c.C° Density of ice, ρ = . g/cc Thickness of ice, L = 5.00 cm REQUIRED: Rate of formation of ice,

dL d

SOLUTION: Solving for H H=

  

------------------ equation 1

Solving for Q  Q = mL f --------------- equation 2 Solving for m m = ρV ---------------- equation 3 Solving for V V = AL ---------------- equation 4 Substitute equation 4 to equation 3 m = ρAL ------------- equation 5 Substitute equation 5 to equation 2  = ρ A L L f ------------- equation 6 Substitute equation 6 to equation 1 H=

ALL

f 



------------- equation 7

Solving for H using conduction formula H=

k A (

 ) 

L

------------- equation 8

Equate equation 7 and equation 8

 A L L = k A (   ) f 





2

L =

L

k (

 ) L  

f 



Darius L. Fajardo

P r o b l e m S e t N o . 1 P M A T 5 0 7 | 14

Evaluate the first derivative of L with respect to T 2L dL =

  ) d L 

k (



f 

dL d dL d L 

=

=

k (





L Lf  ρ

)

. (

()(.)(.)

= 0.197 cm/hr

ANSWER: L

 )

= 0.197 cm/hr

x 3600 cm/hr