What is the heat required in kilocalories to convert 2 kg of ice at 0°C completely into steam at 100°C?
-80 Calories
-1440 Calories
-4186 Calories
-540 Calories
Thanks!
Thanks this answer is very useful and easy to understand
add the heat of fusion, heat of heating, and the heat of vaporization:
heat=2 *80+2 *1*100+ 2*540
= 160+200+1080
add them up
To convert 2 kg of ice at 0°C completely into steam at 100°C, we need to consider the following steps:
1. Heating the ice from 0°C to its melting point (0°C)
2. Melting the ice into water at its melting point
3. Heating the water from its melting point to its boiling point (100°C)
4. Vaporizing the water into steam at its boiling point
To calculate the heat required for each step, we can use the formula:
Q = m * C * ΔT
where:
Q is the heat required
m is the mass of the substance
C is the specific heat capacity
ΔT is the change in temperature
Let's calculate the heat required for each step:
1. Heating the ice from 0°C to its melting point (0°C):
Q1 = 2 kg * 0.5 kcal/kg°C * (0°C - 0°C) = 0 kcal
2. Melting the ice into water at its melting point:
Q2 = 2 kg * 80 kcal/kg = 160 kcal
3. Heating the water from its melting point to its boiling point (100°C):
Q3 = 2 kg * 1 kcal/kg°C * (100°C - 0°C) = 200 kcal
4. Vaporizing the water into steam at its boiling point:
Q4 = 2 kg * 540 kcal/kg = 1080 kcal
Now, let's calculate the total heat required by summing up the heat required for each step:
Total heat required = Q1 + Q2 + Q3 + Q4 = 0 kcal + 160 kcal + 200 kcal + 1080 kcal = 1440 kcal
Therefore, the heat required in kilocalories to convert 2 kg of ice at 0°C completely into steam at 100°C is 1440 kilocalories. Therefore, the correct answer is -1440 Calories.
To calculate the heat required to convert a substance from one phase to another, you can use the formula:
Q = m * L
Where:
Q is the heat required
m is the mass of the substance
L is the specific latent heat of the substance
In this case, we need to convert 2 kg of ice at 0°C to steam at 100°C.
First, to convert ice at 0°C to water at 0°C, we need to provide heat for the phase change from solid to liquid. The specific latent heat of fusion for ice is 80 calories/gram. So, the heat required to melt 2 kg (2000 grams) of ice is:
Q1 = m * L1
= 2000g * 80 cal/g
= 160,000 cal
Next, to raise the temperature of the liquid water from 0°C to 100°C, we need to provide heat to increase its temperature. The specific heat capacity of water is 1 cal/g°C. So, the heat required is:
Q2 = m * c * ΔT
= 2000g * 1 cal/g°C * (100°C - 0°C)
= 200,000 cal
Finally, to convert the water at 100°C to steam at 100°C, we need to provide heat for the phase change from liquid to gas. The specific latent heat of vaporization for water is 540 calories/gram. So, the heat required is:
Q3 = m * L3
= 2000g * 540 cal/g
= 1,080,000 cal
Adding up Q1, Q2, and Q3, we get the total heat required:
Q_total = Q1 + Q2 + Q3
= 160,000 cal + 200,000 cal + 1,080,000 cal
= 1,440,000 cal
= 1,440 kilocalories
Therefore, the correct answer would be 1,440 kilocalories.