Question

A $$0.3\,kg$$ of hot coffee, which is at ${70^ \circ }C$ is poured into a cup of mass $$0.12\,kg$$ . Find the final equilibrium temperature. Take room temperature as $${20^ \circ }C$$ , $${s_{coffee}} = 4080{\text{ }}J/kg - K$$ and $${s_{cup}} = 1020J/kg - K$$
A. $$45.5^\circ C$$
B. $$55.5^\circ C$$
C. $$65.5^\circ C$$
D. None of these

Answer 1

Hint: As the temperature of coffee is higher than temperature of the cup, there will be a sharing of temperature, the heat will be transferred from the coffee to the cup. The resultant final temperature will be lower than the ${70^ \circ }C$ and higher than the room temperature, i.e. $${20^ \circ }C$$. The resultant final temperature of cup and coffee will be equal.

Complete step by step answer:
Let us assume that the equilibrium temperature is $T$. Now, heat lost by coffee will be equal to the heat gained by the cup. Or we can say,
$${m_{coffee}}({S_{coffee}})\Delta {T_{coffee}} = {m_{cup}}({S_{cup}})\Delta {T_{cup}}$$
Where, $m$ is mass, $S$ is entropy and $T$ is temperature.
Now, putting the values and solving it,
$0.3 \times 4080 \times (70 - T) = 0.12 \times 1020 \times (T - 20)$
Now, solving the equation for the value of T,
$4080(70 - T) = 408(T - 20) \\
\Rightarrow 4080(70) - 4080(T) = 408T - 408(20) $
Now collecting same type of factors on one side, we have,
$(4080 + 408)T = 4080(70) + 408(20) \\
\therefore T = {65.45^ \circ }C $

Hence, the equilibrium temperature is - $T = {65.45^ \circ }C$ and the correct option is C) $$65.5^\circ C$$.

Note: The dependence on activity and mass are easily understood. due to the very fact that the (average) mechanical energy of an atom or molecule is proportional to absolutely the temperature, the interior energy of a system is proportional to absolutely the temperature and therefore the number of atoms or molecules. Thanks to the very fact that the transferred heat is adequate to the change within the internal energy, the warmth is proportional to the mass of the substance and also the activity.