Question

An ice cube of mass 10 g at $ - 10^\circ C $ is dropped into a calorimeter containing 20 g water at $ 20^\circ C $ . If the water equivalent of a calorimeter is 10 g. then final temperature of mixture in calorimeter would be
(A) $ 3.2^\circ C $
(B) $ 6.4^\circ C $
(C) $ 9.8^\circ C $
(D) $ 0^\circ C $

Answer 1

Hint: We can recall that latent heat of fusion needs to be considered for ice at zero degree Celsius. Check if the heat lost by water if it cools down to zero degrees would be enough to cool the ice and convert it to water.

Formula used: In this solution we will be using the following formulae;
 $ H = mc\Delta T $ where $ H $ is the heat loss or gained by a body, $ m $ is the mass of the body, $ c $ is the specific heat capacity of the body and $ \Delta T $ is the change in temperature of the body, $ \Delta T = {T_f} - {T_i} $ for heat gained, and $ \Delta T = {T_i} - {T_f} $ for heat loss where $ {T_i} $ is initial temperature and $ {T_f} $ is the final temperature.
 $ {H_l} = mL $ where $ {H_l} $ is the latent heat of fusion, $ L $ is the specific latent heat of fusion, but also often called latent heat of fusion.

Complete Step-by-Step solution:
To solve the above, we need to note that ice at zero degrees needs to gain the latent heat of fusion to become water.
First, for such a question, we assume that the heat lost is equal to heat gained to find the equilibrium temperature. If a particular temperature is assumed, and the heat gained and heat loss does not balance at the temp, then we know it is wrong.
Heat lost or gained is given as
 $ H = mc\Delta T $ where $ H $ is the heat loss or gained by a body, $ m $ is the mass of the body, $ c $ is the specific heat capacity of the body and $ \Delta T $ is the change in temperature of the body, while latent heat is given as
 $ {H_l} = mL $ where $ {H_l} $ is the latent heat of fusion, $ L $ is the specific latent heat of fusion,
In our case, let’s assume that the water lost enough heat to the ice enough to melt all the ice to liquid at 0 degree Celsius. In this case, hence, we have
 $ H = 30 \times 1 \times \left( {20 - 0} \right) = 600cal $ (since specific heat of water is $ 1{\text{ }}cal/g \cdot ^\circ C $
Then heat gained by ice to 0 degree is
 $ H = 10 \times 0.5 \times \left( {0 - \left( { - 10} \right)} \right) = 50cal $ and if the ice melts, latent heat of ice is
 $ {H_l} = 10 \times 80 = 800cal $ (specific latent heat of ice is 80 cal/g)
We see that the sum of the heat gained by ice to raise the temperature to 0 degrees and the latent heat to melt all is greater than the heat loss by the water if it cools to zero. However, the heat is greater than the heat required to raise the ice alone. This signifies that the ice gets to zero degrees Celsius indeed, due to the heat absorbed from the water, but there is not enough heat to melt the entire ice into water.
Hence, the mixture must be at $ 0^\circ C $ .
Then, the correct option is D.

Note:
For clarity, note that although the mass of water was 20 g we used 30 g. This is because we were told that the water equivalent of the calorimeter is 10g. This statement means that the heat capacity (not specific heat capacity) of the calorimeter is the same as the heat capacity of 10 g of water. So we can add 10g to the amount of water we already have since the calorimeter is simply behaving as though an extra 10g of water is present.