Question

0.5 kg of lemon squash at $30^\circ C$ is placed in a refrigerator which can remove heat at an average rate of $30\,J{s^{ - 1}}$. How long will it take to cool the lemon squash to $5^\circ C$? Specific heat capacity of squash$= 4200\,J\,k{g^{ - 1}}{K^{ - 1}}$.
A. 29 min 10 s
B. 58 min 20 s
C. 39 min 10 s
D. None of the above

Answer 1

Use the formula for heat gained or lost by the substance of mass m and specific heat C. we have given the rate of heat loss, therefore, calculate the total time taken by the refrigerator to remove the heat you have calculated at first.

Formula used:
The heat lost or gained by the substance of mass m and specific heat capacity C is given as,
$Q = mc\Delta T$, where, $\Delta T$ is the difference in the temperature of the substance.

Complete step by step answer:
We know that according to principle of Calorimetry, the heat lost by the lemon squash after placed in the refrigerator is,
$Q = mc\Delta T$
Here, m is the mass of lemon squash, c is the specific heat capacity and $\Delta T$ is the difference in the final and initial temperature of lemon squash.
Therefore, we can write,
$Q = mc\left( {{T_i} - {T_f}} \right)$

We substitute 0.5 kg for m, $4200\,J\,k{g^{ - 1}}{K^{ - 1}}$ for c, $30^\circ C$ for ${T_i}$ and $5^\circ C$ for ${T_f}$ in the above equation.
$Q = \left( {0.5} \right)\left( {4200} \right)\left( {30 - 5} \right)$
$\Rightarrow Q = 52500\,J$
This is the total heat removed by the refrigerator in the lemon squash. We have given the rate of heat loss, $30\,J{s^{ - 1}}$.

Therefore, we can calculate the time taken by the refrigerator to remove 52500 J of heat as follows,
$t = \dfrac{{52500\,J}}{{30\,J\,{s^{ - 1}}}}$
$\Rightarrow t = 1750\,s$
$\Rightarrow t = \left( {1750\,s} \right)\left( {\dfrac{{1\,\min }}{{60\,s}}} \right)$
$\therefore t = 29\,\min \,\,\,16\,s$

So, the correct answer is option (A).

Note: While using the formula, $Q = mc\Delta T$, the difference in the temperature is in Kelvin. But since it is the difference in the temperature, the difference will be the same for Celsius temperature and Kelvin temperature. While solving these types of questions relating to heat energy loss or gain, students can directly use the formula$Q = mc\Delta T$.