Question

The temperature of equal masses of three different liquids A,B and C are $ \text{1}{{\text{2}}^{\text{o}}}\text{C, 1}{{\text{9}}^{\text{o}}}\text{C and 2}{{\text{8}}^{\text{o}}}\text{C} $ respectively. The temperature when A and B are mixed is $ \text{1}{{\text{6}}^{\text{o}}}\text{C} $ and when B and C are mixed it is $ \text{2}{{\text{3}}^{\text{o}}}\text{C} $ What should be the temperature when A an C are mixed?

Answer 1

Hint: When two substances are mixed together and both are at two different temperatures, then the substance at the higher temperature will lose its heat to raise the temperature of the other substance. This is because heat always flows from higher to lower temperature. If no heat is lost into the surroundings, then by principle of calorimetry the heat lost by the body at higher temperature will be equal to heat gained by the body at lower temperature. The heat gained or lost by the body is numerically equal to $ \text{ }\!\!\Delta\!\!\text{ H=m}{{\text{C}}_{\text{p}}}\text{ }\!\!\Delta\!\!\text{ T J} $.

Complete step-by-step answer:
When two liquids are mixed together the liquid at the lower temperature will absorb the heat from the liquid at higher temperature.
The amount of heat absorbed or heat gained by the liquid is equal to $ \text{ }\!\!\Delta\!\!\text{ H= m}{{\text{C}}_{\text{L}}}\text{ }\!\!\Delta\!\!\text{ T J}....\text{(1)} $ where m is the mass of the liquid, $ {{\text{C}}_{\text{L}}} $ is the specific heat capacity of any liquid and $ \text{ }\!\!\Delta\!\!\text{ T} $ is the change in temperature of the liquid after absorbing or gaining the heat.
Let the specific heat capacity of liquid A,B and C be $ {{\text{C}}_{\text{A}}}\text{,}{{\text{C}}_{\text{B}}}\text{ and }{{\text{C}}_{\text{C}}} $ respectively. Hence, when liquid A and C are mixed, using equation 1 let us try to find out the ratio of their heat capacity.
It is given that when A and B are mixed together their equilibrium temperature is 16 degree Celsius. Hence using equation 1 we get,
 $ \text{Heat lost=Heat gained} $
 $ \text{m}{{\text{C}}_{\text{B}}}\text{(19-16) = m}{{\text{C}}_{\text{A}}}(12-16) $
 $ \begin{align}
  & \dfrac{\text{(19-16)}}{(12-16)}\text{ = }\dfrac{{{\text{C}}_{\text{A}}}}{{{\text{C}}_{\text{B}}}} \\
 & \dfrac{{{\text{C}}_{\text{A}}}}{{{\text{C}}_{\text{B}}}}=\dfrac{4}{4}=1 \\
\end{align} $
Now let us calculate the ratio of the specific heat capacity when the liquids B and C are mixed.
 $ \text{Heat lost=Heat gained} $
 $ \text{m}{{\text{C}}_{\text{C}}}\text{(28-23) = m}{{\text{C}}_{\text{B}}}(23-19) $
 $ \begin{align}
  & \dfrac{\text{(23-19)}}{(28-23)}\text{ = }\dfrac{{{\text{C}}_{\text{C}}}}{{{\text{C}}_{\text{B}}}} \\
 & \dfrac{{{\text{C}}_{\text{C}}}}{{{\text{C}}_{\text{B}}}}=\dfrac{4}{5} \\
\end{align} $
Now further let us find equilibrium temperature when liquid A and C are mixed. For that let us equate the heat lost by liquid C and heat gained by liquid A.
 $ \text{ m}{{\text{C}}_{\text{A}}}\text{ }\!\!\Delta\!\!\text{ T= m}{{\text{C}}_{\text{C}}}\text{ }\!\!\Delta\!\!\text{ T} $
 $ \text{m}{{\text{C}}_{\text{A}}}\text{(T-12) = m}{{\text{C}}_{\text{C}}}(\text{28-T)}...\text{(2)} $ where T is the equilibrium temperature.
Let us divide the ratios obtained above i.e. $ \dfrac{\dfrac{{{\text{C}}_{\text{A}}}}{{{\text{C}}_{\text{B}}}}}{\dfrac{{{\text{C}}_{\text{C}}}}{{{\text{C}}_{\text{B}}}}}=\dfrac{{{\text{C}}_{\text{A}}}}{{{\text{C}}_{\text{C}}}}=\dfrac{1}{\dfrac{4}{5}}=\dfrac{5}{4} $
On solving equation 2 further we get,
 $ \dfrac{{{\text{C}}_{\text{A}}}}{{{\text{C}}_{\text{C}}}}\text{= }\dfrac{\text{(28-T)}}{\text{(T-12)}} $ but we found the ratio of $ \dfrac{{{\text{C}}_{\text{A}}}}{{{\text{C}}_{\text{C}}}} $ is $ \dfrac{5}{4} $ . Hence substituting in the adjacent equation we get,
 $ \dfrac{5}{4}\text{= }\dfrac{\text{(28-T)}}{\text{(T-12)}} $
 $ \begin{align}
  & \text{5T - 60=112 - 4T} \\
 & \text{9T = 172 hence T is 19}\text{.}{{\text{1}}^{\text{o}}}\text{C} \\
\end{align} $
Hence the answer to the above question is 19.1 degree Celsius.

Note: All the heat is transferred from one liquid to another that is no heat is lost into the surroundings. The heat always flows from higher temperature to lower temperature. It is not that this is not possible. As per the entropy of the system, the probability of heat flowing from lower to higher temperature is so less that in the real world it does not actually take place.