How many calories of heat energy are required to melt 20 grams of ice at 0 °C?
Answer: _____ calories
How many calories of heat are required to evaporate 20 grams of liquid water at 0 °C? Answer: _____ calories
15,300 cal
To determine the number of calories of heat energy required to melt 20 grams of ice at 0 °C, we need to use the specific heat capacity and the heat of fusion of ice.
The specific heat capacity (c) of ice is 0.5 calories per gram per degree Celsius (cal/g°C), and the heat of fusion (ΔHf) of ice is 79.7 calories per gram (cal/g).
The equation to calculate the heat energy required to melt an amount of ice is: Q = m × ΔHf, where Q is the heat energy, m is the mass, and ΔHf is the heat of fusion.
For the given mass of ice (m = 20 grams), we can calculate the heat energy required to melt it as follows:
Q = m × ΔHf
Q = 20 grams × 79.7 cal/g
Q = 1594 calories
Therefore, the number of calories of heat energy required to melt 20 grams of ice at 0 °C is 1594 calories.
Now, let's determine the number of calories of heat energy required to evaporate 20 grams of liquid water at 0 °C. For this, we need to use the specific heat capacity and the heat of vaporization of water.
The specific heat capacity (c) of water is 1 calorie per gram per degree Celsius (cal/g°C), and the heat of vaporization (ΔHv) of water is 540 calories per gram (cal/g).
The equation to calculate the heat energy required to evaporate an amount of liquid water is: Q = m × ΔHv, where Q is the heat energy, m is the mass, and ΔHv is the heat of vaporization.
For the given mass of liquid water (m = 20 grams), we can calculate the heat energy required to evaporate it as follows:
Q = m × ΔHv
Q = 20 grams × 540 cal/g
Q = 10800 calories
Therefore, the number of calories of heat energy required to evaporate 20 grams of liquid water at 0 °C is 10800 calories.
To find how many calories of heat energy are required to melt 20 grams of ice at 0 °C, we need to use the specific heat capacity and heat of fusion of ice.
1. First, we need to calculate the heat required to raise the temperature of the ice from 0 °C to its melting point, which is also 0 °C. The specific heat capacity of ice is 0.5 calories/gram·°C.
Heat required for temperature change = Mass × Specific heat capacity × Change in temperature
= 20 grams × 0.5 calories/gram·°C × (0 °C - 0 °C)
= 0 calories
2. Now, we need to calculate the heat required to melt the ice at its melting point. The heat of fusion of ice is 79.7 calories/gram.
Heat required for melting = Mass × Heat of fusion
= 20 grams × 79.7 calories/gram
= 1594 calories
Therefore, the total heat energy required to melt 20 grams of ice at 0 °C is 1594 calories.
To find how many calories of heat energy are required to evaporate 20 grams of liquid water at 0 °C, we need to use the specific heat capacity and heat of vaporization of water.
1. First, we need to calculate the heat required to raise the temperature of the liquid water from 0 °C to its boiling point, which is 100 °C. The specific heat capacity of water is 1 calorie/gram·°C.
Heat required for temperature change = Mass × Specific heat capacity × Change in temperature
= 20 grams × 1 calorie/gram·°C × (100 °C - 0 °C)
= 2000 calories
2. Next, we need to calculate the heat required to vaporize the water at its boiling point. The heat of vaporization of water is 540 calories/gram.
Heat required for vaporization = Mass × Heat of vaporization
= 20 grams × 540 calories/gram
= 10800 calories
Therefore, the total heat energy required to evaporate 20 grams of liquid water at 0 °C is 10800 calories.