Question

On a hot summer day you and some friends decide you want to cool down your pool. Determine the mass of ice you would need to add to bring the equilibrium temperature of the system to 300K. The pool contains 400,000 L (at a density of 1 kg/L) of water initially at 305K. Assume the ice is at 0°C (273K), the heat capacity of water is 4.2 J/(g*K), and the enthalpy of melting ice is 333 kJ/kg.

Answer 1

Answer: The mass of ice you would need to add to bring the equilibrium temperature of the system to 300 K is $16.14 \times 10^{4}$ kg.

Explanation:

We know that relation between heat energy and specific heat is as follows.

                 q = $m \times S \times \Delta T$

As density of water is 1 kg/L and volume is given as 400,000 L. Therefore, mass of water is as follows.

          Mass of water = Volume × Density

                                  = $400,000 L \times 1 kg/L$

                                  = 400,000 kg

or,                              = $400,000 \times 10^{3}$ g    (as 1 kg = 1000 g)

Specific heat of water is 4.2 J/gm K. Therefore, change in temperature is as follows.

         $\Delta T$ = 305 K - 273 K

                    = 32 K

Now, putting the given values into the above formula and calculate the heat energy as follows.

            q = $m \times S \times \Delta T$

                = $400,000 \times 10^{3} \times 4.2 \times 32 K$

                = $5376 \times 10^{7}$ J

or,            = $5376 \times 10^{4}$ kJ

According to the enthalpy of melting of ice 333 kJ/Kg of energy absorbed by by 1 kg of ice. Hence, mass required to absorb energy of $5376 \times 10^{4}$ kJ  is calculated as follows.

            Mass = $\frac{5376 \times 10^{4} kJ}{333 kJ/Kg} \times 1 kg$

                      = $16.14 \times 10^{4}$ kg

Thus, we can conclude that the mass of ice you would need to add to bring the equilibrium temperature of the system to 300 K is $16.14 \times 10^{4}$ kg.