What is the heat released when 25.0 grams of water freezes at 0°C?

1 Answer
Sep 16, 2016

To calculate the amount of heat entering or leaving a system, the equation #Q=mcΔT# is used.


m = mass (in grams)
c = specific heat capacity (J/g°C)
ΔT = change in temperature (°C)

Here, we will use the specific heat capacity for liquid water which is 4.19 J/g°C.

The mass given is 25.0 grams.

As for the change in temperature, I will assume that it start off at room temperature, 25°C.

#25°C - 0°C = 25°C#


# Q = 25 grams * 4.19 J/(g°C) * 25°C#

#Q = 2618.75 J#

Take into account significant figures and the answer should be

# 2.6 * 10^3 J#