How much heat is released when 10 g of steam at 100°C is cooled to 10 g of liquid water at 100°C?
I am working on a practice test and the answer is not in the back of the book. Can some one help me.
1. 10 cal
2. 5400 cal
3. 800 cal
4. 6200 cal
To determine the amount of heat released when steam is cooled to liquid water, we need to use the formula for heat transfer. The formula is:
Q = m * ΔT * c
Where:
Q is the heat transfer in calories
m is the mass of the substance in grams
ΔT is the change in temperature
c is the specific heat capacity of the substance
In this case, we have:
m = 10 g (mass of steam and liquid water)
ΔT = 100°C - 100°C = 0°C (no change in temperature since both are at 100°C)
c = specific heat capacity of water = 1 cal/g°C (can assume steam has the same specific heat capacity as liquid water)
Using these values, we can calculate the heat transfer using the formula:
Q = 10 g * 0°C * 1 cal/g°C = 0 cal
So, none of the options provided in the question (10 cal, 5400 cal, 800 cal, 6200 cal) are correct. The correct answer is 0 cal.
To determine the heat released when steam is cooled to liquid water, we need to calculate the heat lost during the phase change from steam to water and the heat lost during the temperature decrease from 100°C to 100°C.
First, let's calculate the heat lost during the phase change from steam to water:
The heat released during a phase change can be calculated using the formula:
Q = m * ΔHvap
where Q is the heat released, m is the mass of the substance, and ΔHvap is the heat of vaporization.
For water, the heat of vaporization is typically 540 cal/g. Therefore, the heat lost during the phase change from steam to water is:
Q1 = 10g * 540 cal/g = 5400 cal
Next, let's calculate the heat lost during the temperature decrease from 100°C to 100°C:
The formula to calculate the heat lost during a temperature change is:
Q = m * C * ΔT
where Q is the heat lost, m is the mass of the substance, C is the specific heat capacity, and ΔT is the change in temperature.
For water, the specific heat capacity is approximately 1 cal/g°C. Therefore, the heat lost during the temperature decrease from 100°C to 100°C is:
Q2 = 10g * 1 cal/g°C * (100°C - 100°C) = 0 cal
Finally, we can calculate the total heat released by adding Q1 and Q2:
Q_total = Q1 + Q2 = 5400 cal + 0 cal = 5400 cal
So, the correct answer is option 2. 5400 cal