How many calories of heat are required to convert 230 grams of ice at -10.0C to a gas with a temperature of 1500C?
To calculate the number of calories of heat required to convert 230 grams of ice at -10.0°C to a gas with a temperature of 1500°C, we need to go through three steps:
Step 1: Calculate the energy required to warm the ice from -10.0°C to the melting point (0°C).
Step 2: Calculate the energy required to convert the ice at 0°C to water at 0°C.
Step 3: Calculate the energy required to convert the water at 0°C to a gas at 1500°C.
Step 1: To warm the ice from -10.0°C to the melting point (0°C), we need to use the specific heat capacity of ice:
Q1 = m * c * ΔT
Q1 = 230 g * 2.09 cal/g°C * ((0°C) - (-10.0°C))
Q1 = 230 g * 2.09 cal/g°C * 10.0°C
Q1 = 4814 cal
Step 2: To convert the ice at 0°C to water at 0°C, we need to use the heat of fusion value:
Q2 = m * ΔHf
Q2 = 230 g * 79.7 cal/g
Q2 = 18331 cal
Step 3: To convert the water at 0°C to a gas at 1500°C, we need to use the specific heat capacity of water (liquid phase) and the heat of vaporization value:
Q3 = (m * c * ΔT) + (m * ΔHv)
Q3 = (230 g * 1 cal/g°C * (1500°C - 0°C)) + (230 g * 540 cal/g)
Q3 = 230 g * (1 cal/g°C * 1500°C + 540 cal/g)
Q3 = 391,200 cal
Finally, we can add up the three steps to get the total energy required:
Total Q = Q1 + Q2 + Q3
Total Q = 4814 cal + 18331 cal + 391,200 cal
Total Q ≈ 414,345 cal
Therefore, approximately 414,345 calories of heat are required to convert 230 grams of ice at -10.0°C to a gas with a temperature of 1500°C.
To find the number of calories of heat required to convert ice at -10.0°C to a gas at 1500°C, we need to consider the various phase changes that the substance undergoes.
First, we need to calculate the calories of heat required to raise the temperature of the ice from -10.0°C to its melting point of 0°C.
The specific heat capacity of ice is approximately 0.5 calories/gram·°C. So, the heat required would be:
Q1 = mass × specific heat capacity × change in temperature
= 230 g × 0.5 cal/g·°C × (0°C - (-10.0°C))
= 230 g × 0.5 cal/g·°C × 10.0°C
= 1150 cal
Next, we need to calculate the calories of heat required to melt the ice at 0°C to water at 0°C. The heat of fusion for ice is approximately 80 calories/gram. So, the heat required would be:
Q2 = mass × heat of fusion
= 230 g × 80 cal/g
= 18,400 cal
Now, we need to calculate the calories of heat required to raise the temperature of the water from 0°C to its boiling point at 100°C. The specific heat capacity of liquid water is approximately 1.0 calorie/gram·°C. So, the heat required would be:
Q3 = mass × specific heat capacity × change in temperature
= 230 g × 1 cal/g·°C × (100°C - 0°C)
= 230 g × 1 cal/g·°C × 100°C
= 23,000 cal
Next, we need to calculate the calories of heat required to convert the water at 100°C to steam at 100°C. The heat of vaporization for water is approximately 540 calories/gram. So, the heat required would be:
Q4 = mass × heat of vaporization
= 230 g × 540 cal/g
= 124,200 cal
Finally, we need to calculate the calories of heat required to raise the temperature of the steam from 100°C to 1500°C. The specific heat capacity of steam is approximately 0.5 calories/gram·°C. So, the heat required would be:
Q5 = mass × specific heat capacity × change in temperature
= 230 g × 0.5 cal/g·°C × (1500°C - 100°C)
= 230 g × 0.5 cal/g·°C × 1400°C
= 161,000 cal
Now, we can add up all the heat values:
Total heat required = Q1 + Q2 + Q3 + Q4 + Q5
= 1150 cal + 18,400 cal + 23,000 cal + 124,200 cal + 161,000 cal
= 327,750 cal
Therefore, approximately 327,750 calories of heat are required to convert 230 grams of ice at -10.0°C to a gas with a temperature of 1500°C.