Calculate the amount of heat released when 65.6 g of water at 55.5°C cools to ice at 0.0°C.
To calculate the amount of heat released when water cools, we can use the formula:
q = m * c * ΔT
where:
q is the heat released (in Joules),
m is the mass of the substance (in grams),
c is the specific heat capacity of the substance, and
ΔT is the change in temperature.
In this case, we need to consider two temperature changes: from 55.5°C to 0.0°C, and then from 0.0°C to the freezing point of water.
Step 1: Calculate the heat released when water cools from 55.5°C to 0.0°C.
The specific heat capacity (c) of water is approximately 4.18 J/g°C.
q1 = m * c * ΔT1
= 65.6 g * 4.18 J/g°C * (0.0°C - 55.5°C)
= -153033.6 J
The negative sign indicates that heat is being released.
Step 2: Calculate the heat released when water freezes to ice.
The specific heat capacity of ice is approximately 2.09 J/g°C.
q2 = m * c * ΔT2
= 65.6 g * 2.09 J/g°C * (0.0°C - (-0.0°C))
= 0 J
Since the temperature doesn't change during phase change, no heat is released.
Step 3: Calculate the total heat released.
The total heat released (q) is the sum of q1 and q2.
q = q1 + q2
= -153033.6 J + 0 J
= -153033.6 J
Therefore, the amount of heat released when 65.6 g of water at 55.5°C cools to ice at 0.0°C is approximately -153033.6 Joules.