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I will upload sheet for assignment andpicturesfor questions , please answer the fellowing questions with details :4-3, 4-4, 4-8, 4-12, 4-15, 2, 3 .Classical and Statistical Thermodynamics, Carter Chapter 4 – 4.3 and 4.4 – 4.8 and 4.12 – XXXXXXXXXXA 25 kg cast-iron (Ciron = 0.42) wood burning stove contains 5 kg of soft pine wood (Cwood = 1.38) and 1 kg of air (Cair = XXXXXXXXXXAll the masses are at room temperature, 20 oC, and pressure 101 kP a. The wood now burns and heats all the mass uniformly, releasing 1500 W. Neglect any air flow and changes in mass of wood and heat losses. Find the rate of change of temperature (dT/dt) and estimate the time it will take to reach a temperature of 75oC. 3) During the charging of a storage battery, the current i is 20 A and the voltage ? is 12.8 V . The rate of heat transfer from the battery is 10 W. At what rate is the thermal energy increasing?
	
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Physics 306
Homework Assignment 4
(Heat Capacities and Enthalpy.)
1) Classical and Statistical Thermodynamics, Carter Chapter 4
– 4.3 and 4.4
– 4.8 and 4.12
– 4.15
2) A 25kg cast-iron (C = 0.42) wood burning stove contains 5kg of soft pine wood
iron
(C = 1.38) and 1kg of air (C = XXXXXXXXXXAll the masses are at room temperature,
wood air
o
20 C, and pressure 101kPa. The wood now burns and heats all the mass uniformly,
releasing 1500 W. Neglect any air ?ow and changes in mass of wood and heat losses.
Find the rate of change of temperature (dT/dt) and estimate the time it will take to
o
reach a temperature of 75 C.
3) During the charging of a storage battery, the current i is 20A and the voltage ? is
12.8V. The rate of heat transfer from the battery is 10W. At what rate is the
thermal energy increasing?

Robert · Accepted Answer

2) Given Data: 
Mass of iron ‘ miron’ = 25 kg 
Mass of wood ‘mwood’ = 5 kg 
Mass of air ‘mair’ = 1 kg 
Ciron = 0.42 
Cwood = 1.38 
Cair = 0.717 
Asuumptions: 
i) There are no changes in the kinetic and potential energy. 
ii) There is no work-done. 
iii) It’s already given no changes in mass. 
Now using the 1st law of thermodynamics we have  
Q = mU+W 
As W = 0, Therefore 
1500 = (miron*Ciron+ mwood*Cwood+ mair*Cair)*(dT/dt) 
So, dT/dt = 1500/(25*0.42+5*1.38+1*0.717) 
dT/dt = 0.0828 K/s (Ans)
Now, time to reach 750C from 200C 
t = (75-20)/(dT/dt) = 55/0.0828 = 664.25 sec  (Ans)
3) Given data: 
Current I = 20 A 
Voltage V = 12.8 V 
Rate of heat transfer Q/t =- 10 W (-ve sign because heat is going out from the system) 
Now, neglecting the changes in kinetic and potential energies. We can write the 1st law of 
thermodynamics as  
Q/t = (dU/dt) +W/t 
Now W/t = -(V*i) = -(12.8*20) = -256 W  (-ve sign because the work will be done on the system) 
So, -10 = (dU/dt) + (-256) 
dU/dt = 246 W
4.3) Given eq. 2.13   β = (1/v)*(δv/δT)P
Now we have to prove that (δu/δT)P = CP – Pβv 
Now suing the first law of thermodynamics 
du = Tds _Pdv 
Differentiating both side w.r.

I will upload sheet for assignment andpicturesfor questions , please answer the fellowing questions with details :4-3, 4-4, 4-8, 4-12, 4-15, 2, 3 . Classical and Statistical Thermodynamics, Carter...

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