Question

Liquids A and B are $ 60^\circ $ and $ 20^\circ $ . When mixed in equal masses, the temperature of the mixture is found to be $ 30^\circ $ . Find the ratio of the specific heats of A and B.

Answer 1

Hint : Generally, for in an isolated system, the heat lost by one substance in it would be equal to the heat gained by the other substance (or substances). The final temperature is the temperature of the mixture.

Formula used: In this solution we will be using the following formulae;
 $ Q = mc\Delta T $ where $ Q $ is the heat loss or gain by a substance, $ m $ is the mass of the substance, $ c $ is the specific heat capacity (also known as specific heat) and $ \Delta T $ is the change in temperature after the said amount of heat is lost or gained, and is given by $ \Delta T = {T_f} - {T_i} $ for heat gained and $ \Delta T = {T_i} - {T_f} $ for heat loss, (just to keep it positive) where $ {T_f} $ is the final temperature and $ {T_i} $ is the initial temperature.

Complete step by step answer:
Liquids A and B are said to be at two different temperatures, but they mixed with equal masses and the final temperature was somewhere in between the two temperatures. The ratio of their heat capacity is to be determined.
To solve this, the first thing we do is to consider the mixture as a system with two substances (sub systems). From conservation of energy, we know that the heat lost by one substance must be equal to the heat gained by the other substance. But the heat lost or gained by a substance is given as
 $ Q = mc\Delta T $ where $ Q $ is the heat loss or gain by a substance, $ m $ is the mass of the substance, $ c $ is the specific heat capacity and $ \Delta T $ is the change in temperature after the said amount of heat is lost or gained.
Hence, for liquid A heat is lost, hence
 $ {Q_A} = mc\Delta T = mc(60 - 30) = 30m{c_A} $
For liquid B
 $ {Q_B} = mc\Delta T = mc(30 - 20) = 10m{c_B} $
Hence,
 $ 30m{c_A} = 10m{c_B} $
 $ \dfrac{{{c_A}}}{{{c_B}}} = \dfrac{{10}}{{30}} = \dfrac{1}{3} $
Hence, the ratio is 1:3.

Note:
In the above, we assumed the system to be isolated i.e. the system does not interact with its surroundings and hence does not lose heat to the atmosphere or the container itself, in actuality no system is completely isolated, hence, the some of the heat would have being lost to the surroundings and the heat lost by one substance will be heat lost to the environment plus heat gained by the other substance.