Question

If 1 kg of water at 20 c and 1 kg of ice at -20 c are mixed. what will the final temperature?

Answer 1

Answer on the question #46324, Physics, Molecular Physics | Thermodynamics

Question:

If 1 kg of water at 20 c and 1 kg of ice at -20 c are mixed. what will the final temperature?

Answer:

The processes that take place in the system are:

1. Warming of the ice

2. Cooling of the water

3. Melting of the ice (probably, we need to calculate to elucidate it)

According to the heat capacity definition:

$$Q = \int_{1}^{2} C dT$$

Where 1 and 2 are initial and finish conditions of the system and C is heat capacity for the substance.

Let's write the equations for heating ice and cooling water:

$$\begin{array}{l} Q = C_{ice} * m_{ice}(0 - T_{1ice}) = 2.11 * 1000 * (0 + 20) = 4.22 * 10^{4} J \\ - Q = C_{water} * m_{water}(T_{2water} - T_{1water}), \end{array}$$

Then, $T_{2water}$ is:

$$T_{2water} = T_{1water} - \frac{Q}{C_{water} * m_{water}} = 20 - \frac{4.22 * 10^{4}}{4.1813 * 10^{3}} = 20 - 10.09 = 9.91 \, {}^{\circ}\mathrm{C}$$

As $T_{2water}$ is $>0 \, {}^{\circ}\mathrm{C}$, the part of the ice melts.

$$\begin{array}{l} - Q = C_{water} * m_{water}(0 - 9.91) = -41.42 * 10^{3} J \\ Q = \Delta H_{fus} * m_{ice} \\ m_{ice.melt} = \frac{Q}{\Delta H_{fus}} = \frac{41.42 * 1000}{333.55} = 124.2 \, g \\ \end{array}$$

Then, the part (57 g) of ice melts. The water and the ice are in equilibrium at $0{}^{\circ}\mathrm{C}$.

http://www.AssignmentExpert.com/