calories to freeze 35g of water
calories to freeze 250g of water
kilocalories to melt 140g of ice
To heat water:
mass H2O x specific heat H2O x (Tfinal-Tinital)
T melt ice:
q = mass ice x heat fusion water
.140kc
The amount of energy required to freeze or melt a substance can be calculated using the specific heat capacity and latent heat of fusion/vaporization of the substance.
For water:
- The specific heat capacity of water is 1 calorie/gram °C.
- The latent heat of fusion of water is 80 calories/gram.
- The latent heat of vaporization of water is 540 calories/gram.
To freeze water:
1. Calculate the energy required to cool the water from its initial temperature to its freezing point (0°C):
Energy = mass * specific heat capacity * temperature difference
Energy = 35g * 1 cal/g°C * (0°C - initial temperature)
2. Calculate the energy required to freeze the water at 0°C:
Energy = mass * latent heat of fusion
Energy = 35g * 80 cal/g
To melt ice:
1. Calculate the energy required to melt the ice at 0°C:
Energy = mass * latent heat of fusion
Energy = 140g * 80 cal/g
Note: 1 kilocalorie (kcal) is equal to 1000 calories.
To convert the energy calculated above into kilocalories, divide the energy value by 1000:
Energy in kilocalories = Energy in calories / 1000
Therefore, to answer your questions:
1. Calories to freeze 35g of water:
Energy = (35g * 1 cal/g°C * (0°C - initial temperature)) + (35g * 80 cal/g)
Energy in kilocalories = Energy in calories / 1000
2. Calories to freeze 250g of water:
Energy = (250g * 1 cal/g°C * (0°C - initial temperature)) + (250g * 80 cal/g)
Energy in kilocalories = Energy in calories / 1000
3. Kilocalories to melt 140g of ice:
Energy = 140g * 80 cal/g
Energy in kilocalories = Energy in calories / 1000
To calculate the calories required for a phase change, such as freezing or melting, you need to know the specific heat capacity and the heat of fusion of the substance.
For freezing, the specific heat capacity of water is approximately 1 calorie/gram °C, and the heat of fusion is approximately 79.7 calories/gram. This means that to freeze each gram of water, it requires 79.7 calories to reach the freezing point, and an additional 1 calorie to freeze it.
To find the calories required to freeze a certain amount of water:
1) For 35g of water:
- Multiply the mass (35g) by the specific heat capacity (1 calorie/gram °C) to determine the calories required to reach the freezing point: 35g * 1 calorie/gram °C = 35 calories.
- Add the heat of fusion (79.7 calories/gram) to freeze the water: 35 calories + 79.7 calories = 114.7 calories.
Therefore, it would require approximately 114.7 calories to freeze 35g of water.
2) For 250g of water:
- Multiply the mass (250g) by the specific heat capacity (1 calorie/gram °C) to determine the calories required to reach the freezing point: 250g * 1 calorie/gram °C = 250 calories.
- Add the heat of fusion (79.7 calories/gram) to freeze the water: 250 calories + 79.7 calories = 329.7 calories.
Therefore, it would require approximately 329.7 calories to freeze 250g of water.
Now, for melting, the process is slightly different. To melt ice, you need to consider the heat of fusion, which is the same as the heat of freezing (79.7 calories/gram) since it is an opposite phase change.
To find the kilocalories required to melt a certain amount of ice:
1) For 140g of ice:
- Multiply the mass (140g) by the heat of fusion (79.7 calories/gram) to determine the calories required to melt the ice: 140g * 79.7 calories/gram = 11,158 calories.
- Convert calories to kilocalories by dividing by 1000: 11,158 calories / 1000 = 11.158 kilocalories.
Therefore, it would require approximately 11.158 kilocalories to melt 140g of ice.