Question
1.
How many joules are needed to vaporize 125 g of liquid water at 100°C? The latent heat of vaporization is 2258 J/g.
282,000 J
2.82 J
2,820,000 J
282 J
2. What requires the most energy?
Raising the temperature by 1 degree of 5 kg of ice
Freezing 0.5 kg of ice
Melting 1 kg of ice
Raising the temperature by 1 degree for 1 kg of ice
3. An unknown substance requires 32340 J of heat to melt. If the substance has a mass of 1.32 kg, what is the unknown substance?
Substance Lf (J/kg) Lv (J/kg) Melting Point (°C) Boiling Point (°C)
water 3.33 x 105 2.26 x 106 0.00 100.00
lead 2.45 x 104 8.70 x 105 327.30 1750.00
gold 6.44 x 104 1.58 x 106 1063.00 2660.00
aluminum 3.97 x 105 1.14 x 107 660.00 2450.00
aluminum
lead
gold
water
4.
If a substance is changing temperature from below its melting point to above its melting point, how many calculations would you have to perform?
1
3
5
4
5.
You have 1 kg of water and you want to use that to melt 0.1 kg of ice. What is the minimum temperature necessary in the water, to just barely melt all of the ice? (Lf = 3.33 x 105 J/kg, cwater 4186 J/kg°C)
8°C
100°C
80°C
18°C
6.
During a phase change, what properties change?
Volume
Temperature
Mass
Heat
7. A piece of ice is melting on a table. How does the temperature just before melting compare to the temperature just after melting?
impossible to tell
before = after
before > after
before < after
8.
Aluminum is often melted down for recycling purposes. Assuming 1 kg of aluminum is at 20 deg C (room temperature), what is the minimum amount of heat needed to melt it to a liquid? The specific heat capacity of aluminum is 899 J/kg°C, the melting point is 660°C, and the latent heat of fusion is 3.97 x 105 J/kg.
571,000 J
397,000 J
972,000 J
500,000 J
9. How many kilograms of water at 100.0°C can be vaporized if 1278900 J of heat are added to the water? (Lv = 2.26 x 106 J/kg)
0.987 kg
0.566 kg
0.653
0.342 kg
10. Which process generally requires more energy?
Fusion
Melting
Heating a substance by one degree
Vaporization
How many joules are needed to vaporize 125 g of liquid water at 100°C? The latent heat of vaporization is 2258 J/g.
282,000 J
2.82 J
2,820,000 J
282 J
2. What requires the most energy?
Raising the temperature by 1 degree of 5 kg of ice
Freezing 0.5 kg of ice
Melting 1 kg of ice
Raising the temperature by 1 degree for 1 kg of ice
3. An unknown substance requires 32340 J of heat to melt. If the substance has a mass of 1.32 kg, what is the unknown substance?
Substance Lf (J/kg) Lv (J/kg) Melting Point (°C) Boiling Point (°C)
water 3.33 x 105 2.26 x 106 0.00 100.00
lead 2.45 x 104 8.70 x 105 327.30 1750.00
gold 6.44 x 104 1.58 x 106 1063.00 2660.00
aluminum 3.97 x 105 1.14 x 107 660.00 2450.00
aluminum
lead
gold
water
4.
If a substance is changing temperature from below its melting point to above its melting point, how many calculations would you have to perform?
1
3
5
4
5.
You have 1 kg of water and you want to use that to melt 0.1 kg of ice. What is the minimum temperature necessary in the water, to just barely melt all of the ice? (Lf = 3.33 x 105 J/kg, cwater 4186 J/kg°C)
8°C
100°C
80°C
18°C
6.
During a phase change, what properties change?
Volume
Temperature
Mass
Heat
7. A piece of ice is melting on a table. How does the temperature just before melting compare to the temperature just after melting?
impossible to tell
before = after
before > after
before < after
8.
Aluminum is often melted down for recycling purposes. Assuming 1 kg of aluminum is at 20 deg C (room temperature), what is the minimum amount of heat needed to melt it to a liquid? The specific heat capacity of aluminum is 899 J/kg°C, the melting point is 660°C, and the latent heat of fusion is 3.97 x 105 J/kg.
571,000 J
397,000 J
972,000 J
500,000 J
9. How many kilograms of water at 100.0°C can be vaporized if 1278900 J of heat are added to the water? (Lv = 2.26 x 106 J/kg)
0.987 kg
0.566 kg
0.653
0.342 kg
10. Which process generally requires more energy?
Fusion
Melting
Heating a substance by one degree
Vaporization
Answers
GPT-4o mini
1. To calculate the energy required to vaporize 125 g of water at 100°C using the latent heat of vaporization \( L_v = 2258 \, \text{J/g} \):
\[
Q = m \cdot L_v = 125 \, \text{g} \times 2258 \, \text{J/g} = 282250 \, \text{J}
\]
So, the answer is approximately **282,000 J**.
---
2. To determine which process requires the most energy, we can analyze the processes:
- Raising the temperature by 1 degree of 5 kg of ice requires \( 5 \, \text{kg} \times c_{ice} \times 1 \, °C \).
- Freezing 0.5 kg of ice uses \( L_f \times 0.5 \, \text{kg} \).
- Melting 1 kg of ice uses \( L_f \times 1 \, \text{kg} \).
- Raising the temperature by 1 degree for 1 kg of ice requires \( c_{ice} \times 1 \, °C \).
Since \( L_f \) (latent heat of fusion for ice) is significantly greater than the specific heat capacities, melting 1 kg of ice requires the most energy. The answer is **Melting 1 kg of ice**.
---
3. To find the latent heat of fusion (\( L_f \)) for the unknown substance:
\[
L_f = \frac{Q}{m} = \frac{32340 \, \text{J}}{1.32 \, \text{kg}} \approx 24409.09 \, \text{J/kg}
\]
Comparing this value with the given latent heats:
- Water: \( 3.33 \times 10^5 \, \text{J/kg} \)
- Lead: \( 2.45 \times 10^4 \, \text{J/kg} \)
- Gold: \( 6.44 \times 10^4 \, \text{J/kg} \)
- Aluminum: \( 3.97 \times 10^5 \, \text{J/kg} \)
It seems closest to **lead**.
---
4. If a substance is changing temperature from below its melting point to above its melting point, you need to consider:
1. Heating the solid to its melting point.
2. Melting the solid to a liquid.
3. Heating the liquid to a final temperature above the melting point.
So, the answer is **3** calculations.
---
5. To find the minimum temperature needed in the water to melt 0.1 kg of ice, we calculate the heat needed to melt the ice:
\[
Q = L_f \cdot m_{ice} = 3.33 \times 10^5 \, \text{J/kg} \cdot 0.1 \, \text{kg} = 33300 \, \text{J}
\]
Now, calculate the minimum temperature of water:
\[
Q_{water} = m_{water} \cdot c_{water} \cdot \Delta T = 1 \, \text{kg} \cdot 4186 \, \text{J/kg°C} \cdot (T - 0)
\]
Setting the heat exchanged equal:
\[
33300 = 1 \times 4186 \times (T - 0)
\]
\[
T = \frac{33300}{4186} = 7.94 \, °C \approx 8 \, °C
\]
So, the answer is **8°C**.
---
6. During a phase change, **Temperature** remains constant, but:
- **Volume** can change.
- **Mass** does not change.
- **Heat** is absorbed or released.
The answer is **Volume** can change.
---
7. The temperature just before melting and just after melting is constant (at 0°C for ice). Thus:
The answer is **before = after**.
---
8. To calculate the heat needed to melt 1 kg of aluminum starting at 20°C:
1. Heating aluminum to melting point:
\[
Q_1 = m \cdot c_{Al} \cdot \Delta T = 1 \cdot 899 \cdot (660 - 20) = 1 \cdot 899 \cdot 640 \approx 575360 \, \text{J}
\]
2. Latent heat required to melt the aluminum:
\[
Q_2 = L_f \cdot m = 3.97 \times 10^5 \cdot 1 = 397000 \, \text{J}
\]
Total heat:
\[
Q_{total} = Q_1 + Q_2 = 575360 + 397000 \approx 972360 \, \text{J} \approx 972000 \, \text{J}
\]
So, the answer is **972,000 J**.
---
9. To find the kilograms of water vaporized by \( 1278900 \, \text{J} \):
\[
m = \frac{Q}{L_v} = \frac{1278900 \, \text{J}}{2.26 \times 10^6 \, \text{J/kg}} \approx 0.566 \, \text{kg}
\]
So, the answer is **0.566 kg**.
---
10. The process that generally requires more energy is **Vaporization** compared to fusion, melting, or heating a substance by one degree.
\[
Q = m \cdot L_v = 125 \, \text{g} \times 2258 \, \text{J/g} = 282250 \, \text{J}
\]
So, the answer is approximately **282,000 J**.
---
2. To determine which process requires the most energy, we can analyze the processes:
- Raising the temperature by 1 degree of 5 kg of ice requires \( 5 \, \text{kg} \times c_{ice} \times 1 \, °C \).
- Freezing 0.5 kg of ice uses \( L_f \times 0.5 \, \text{kg} \).
- Melting 1 kg of ice uses \( L_f \times 1 \, \text{kg} \).
- Raising the temperature by 1 degree for 1 kg of ice requires \( c_{ice} \times 1 \, °C \).
Since \( L_f \) (latent heat of fusion for ice) is significantly greater than the specific heat capacities, melting 1 kg of ice requires the most energy. The answer is **Melting 1 kg of ice**.
---
3. To find the latent heat of fusion (\( L_f \)) for the unknown substance:
\[
L_f = \frac{Q}{m} = \frac{32340 \, \text{J}}{1.32 \, \text{kg}} \approx 24409.09 \, \text{J/kg}
\]
Comparing this value with the given latent heats:
- Water: \( 3.33 \times 10^5 \, \text{J/kg} \)
- Lead: \( 2.45 \times 10^4 \, \text{J/kg} \)
- Gold: \( 6.44 \times 10^4 \, \text{J/kg} \)
- Aluminum: \( 3.97 \times 10^5 \, \text{J/kg} \)
It seems closest to **lead**.
---
4. If a substance is changing temperature from below its melting point to above its melting point, you need to consider:
1. Heating the solid to its melting point.
2. Melting the solid to a liquid.
3. Heating the liquid to a final temperature above the melting point.
So, the answer is **3** calculations.
---
5. To find the minimum temperature needed in the water to melt 0.1 kg of ice, we calculate the heat needed to melt the ice:
\[
Q = L_f \cdot m_{ice} = 3.33 \times 10^5 \, \text{J/kg} \cdot 0.1 \, \text{kg} = 33300 \, \text{J}
\]
Now, calculate the minimum temperature of water:
\[
Q_{water} = m_{water} \cdot c_{water} \cdot \Delta T = 1 \, \text{kg} \cdot 4186 \, \text{J/kg°C} \cdot (T - 0)
\]
Setting the heat exchanged equal:
\[
33300 = 1 \times 4186 \times (T - 0)
\]
\[
T = \frac{33300}{4186} = 7.94 \, °C \approx 8 \, °C
\]
So, the answer is **8°C**.
---
6. During a phase change, **Temperature** remains constant, but:
- **Volume** can change.
- **Mass** does not change.
- **Heat** is absorbed or released.
The answer is **Volume** can change.
---
7. The temperature just before melting and just after melting is constant (at 0°C for ice). Thus:
The answer is **before = after**.
---
8. To calculate the heat needed to melt 1 kg of aluminum starting at 20°C:
1. Heating aluminum to melting point:
\[
Q_1 = m \cdot c_{Al} \cdot \Delta T = 1 \cdot 899 \cdot (660 - 20) = 1 \cdot 899 \cdot 640 \approx 575360 \, \text{J}
\]
2. Latent heat required to melt the aluminum:
\[
Q_2 = L_f \cdot m = 3.97 \times 10^5 \cdot 1 = 397000 \, \text{J}
\]
Total heat:
\[
Q_{total} = Q_1 + Q_2 = 575360 + 397000 \approx 972360 \, \text{J} \approx 972000 \, \text{J}
\]
So, the answer is **972,000 J**.
---
9. To find the kilograms of water vaporized by \( 1278900 \, \text{J} \):
\[
m = \frac{Q}{L_v} = \frac{1278900 \, \text{J}}{2.26 \times 10^6 \, \text{J/kg}} \approx 0.566 \, \text{kg}
\]
So, the answer is **0.566 kg**.
---
10. The process that generally requires more energy is **Vaporization** compared to fusion, melting, or heating a substance by one degree.