Question

The temperature of equal masses of three different liquids A, B and C are 12, 19 and 28 respectively. The temperature when A and B is mixed is 16 and when B and C are mixed is 23. The temperature when A and C is mixed is:
a.) 18.2
b.) 22
c.) 20.2
d.) 24.2

Answer 1

Hint: The heat absorbed by a substance is equal to the mass of the substance multiplied by its specific heat and the change in temperature. The specific heat of a substance is the amount of heat per unit mass required to raise the temperature of the same substance by one degree Celsius.

Complete step by step answer:
Specific heat capacity is the amount of energy that must be added, in the form of heat, to one unit of mass of the substance in order to cause an increase of one unit in its temperature.
Now, in the given question, let the mass of each liquid be m and the specific heat capacities of A, B and C be \[{S_A},\,{S_B},\,{S_C}\] respectively.
When A and B are mixed, the final temperature of the mixture is 16.
Thus, heat gained by A = heat lost by B
\[ \Rightarrow m{S_A}(16 - 12) = m{S_B}(19 - 16)\]
\[ \Rightarrow {S_B} = \dfrac{4}{3}{S_A}\]… eq. 1
And similarly,
\[{S_C} = \dfrac{4}{5}{S_B}\]… eq. 2

Using eq. 1 in eq. 2, we get,
\[{S_C} = \dfrac{4}{5} \times \dfrac{4}{3}{S_A}\]
\[ \Rightarrow {S_C} = \dfrac{{16}}{{15}}{S_A}\]
Now, when A and C are mixed, let the final temperature be T
\[\therefore m{S_A}(T - 12) = m{S_C}(28 - T)\]
\[ \Rightarrow T - 12 = \dfrac{{16}}{{15}}(28 - T)\]
Solving the equation, we get
T = 20.2

Hence, the correct answer is (C)

Note: Remember that the specific heat capacity of a substance is also equal to the heat capacity of a sample of the substance divided by the mass of the sample.