Question

How much energy is needed to cool a 250g cup of water from 23 °C to -23 °C?

Answer 1

During the cooling process, we can identify three stages, as shown in the figure attached to the question.

1. Water cooling from 23°C to 0°C, freezing temperature of the water. The energy for this stage will be:

$Q_1=mCp_1\Delta T$

Where,

m is the mass of water, 250g

Cp1 is the specif heat of water, 4.186J/g°C

dT is the difference of temperature, T2-T1=0°C-23°C=-23°C

$Q_1=250g\times4.186\frac{J}{g\degree C}\times-23\degree C=-24069.5J$

We have a negative value because the process releases energy

2. Change of phase of water, from liquid water to ice. This process occurs at a constant temperature equal to 0°C. The energy for this stage will be:

$Q_2=-m\Delta Hf$

dHf is the heat of fusion. We put a negative sign because we have the contrary process of fusion, freezing.

$\begin{gathered} Q_2=250g\times-334J/g \\ Q_2=-83500J \end{gathered}$

3. Ice cooling from 0°C to -23°C.

$Q_3=mCp_3\Delta T$

Where,

m is the mass of ice, 250g

Cp3 is the specific heat of ice, 2.1J/g°C

dT is the difference of temperature, -23°C-0°C=-23°C

$\begin{gathered} Q_3=250g\times2.1\frac{J}{g\degree C}\times-23\degree C \\ Q_3=-12075J \end{gathered}$

So, the total energy released will be:

$\begin{gathered} Q_T=Q_1+Q_2+Q_3 \\ Q_T=119644.5J=-119.6kJ \end{gathered}$

Answer: The total energy will be -119.6kJ, which means that the energy is released.