Examples of calculations asked on heat and thermodynamics with answers in physics 121?

1. Calculate the amount of heat required to raise the temperature of 500 grams of water from 20°C to 50°C. (Specific heat capacity of water = 4.18 J/g°C)

Q = mcΔT
Q = (500g)(4.18 J/g°C)(50°C - 20°C)
Q = (500)(4.18)(30)
Q = 6270 J

2. A 2 kg block of iron at 100°C is placed in a bucket containing 5 kg of water at 20°C. Calculate the final temperature of the system assuming no heat is lost to the surroundings. (Specific heat capacity of iron = 0.45 J/g°C, specific heat capacity of water = 4.18 J/g°C)

Qlost = Qgain
m1c1ΔT1 = m2c2ΔT2
(2kg)(0.45 J/g°C)(Tf - 100°C) = (5kg)(4.18 J/g°C)(Tf - 20°C)
0.9Tf - 90 = 20.9Tf - 418
418 - 90 = 20.9Tf - 0.9Tf
328 = 20Tf
Tf = 16.4°C

Therefore, the final temperature of the system is 16.4°C.

Example 1:

Question: A 100 g block of ice at -10°C is heated to 20°C. Calculate the amount of heat required to melt the ice and raise its temperature to 20°C.

Answer:

- Heat required to melt the ice: Q = mL = (100 g)(334 J/g) = 33,400 J
- Heat required to raise the temperature of liquid water: Q = mc(T2 - T1) = (100 g)(4.18 J/(g°C))(20°C - 0°C) = 8,360 J
- Total heat required: Q = 33,400 J + 8,360 J = 41,760 J

Example 2:

Question: A 500 L container is filled with helium at 10 atm and 20°C. The temperature is increased to 40°C. Calculate the final pressure of the helium.

Answer:

- Using the Ideal Gas Law:
```
PV = nRT
```
- Assuming the number of moles and volume are constant, we have:
```
P1/T1 = P2/T2
```
- Rearranging for P2, we get:
```
P2 = P1 - T2 / T1 = (10 atm) - (40°C + 273 K) / (20°C + 273 K) = 11.06 atm
```

Example 3:

Question: A heat engine operates at a maximum temperature of 600 K and an exhaust temperature of 300 K. What is the maximum possible efficiency of this engine?

Answer:

- Using the Carnot Efficiency Formula:
```
e = 1 - T2 / T1
```
- Plugging in the values, we get:
```
e = 1 - 300 K / 600 K = 50%
```

Therefore, the maximum possible efficiency of the heat engine is 50%.