The heat of fusion for water is 80cal/g. The specific heat of liquid water is 1.00cal/g. How many calories of heat are released when 20g of water at 30 degrees is cooled to 0 degrees and frozen to ice at 0 degrees?
q1 = heat released when cooled from 30 to zero C.
q1 = mass x specific heat x delta T.
q2 = heat released when water at zero C is frozen.
q2 = mass water x heat fusion.
total heat released is q1 + q2.
To find the total number of calories of heat released, we can break down the process into two steps:
Step 1: Cooling the water from 30°C to 0°C
Step 2: Freezing the water at 0°C to ice at 0°C
Let's calculate the heat released in each step:
Step 1: Cooling the water from 30°C to 0°C
To find the heat released during this step, we need to use the specific heat capacity formula:
Q = m * C * ΔT
where:
Q = heat (in calories)
m = mass of water (in grams)
C = specific heat capacity of water (in cal/g°C)
ΔT = change in temperature (in °C)
In this case:
m = 20g
C = 1.00 cal/g°C
ΔT = (0°C - 30°C) = -30°C
Q1 = (20g) * (1.00 cal/g°C) * (-30°C)
Since the water is cooling down, the heat is being released, so Q1 will be negative.
Q1 = -600 cal
Step 2: Freezing the water at 0°C to ice at 0°C
To find the heat released during this step, we need to use the heat of fusion formula:
Q = m * Hf
where:
Q = heat (in calories)
m = mass of water (in grams)
Hf = heat of fusion of water (in cal/g)
In this case:
m = 20g
Hf = 80 cal/g
Q2 = (20g) * (80 cal/g)
Q2 = 1600 cal
To find the total heat released, we add Q1 and Q2 together:
Total heat released = Q1 + Q2
Total heat released = -600 cal + 1600 cal
Total heat released = 1000 cal
Therefore, when 20g of water at 30 degrees is cooled to 0 degrees and frozen to ice at 0 degrees, the total amount of heat released is 1000 calories.