The temperatures of two liquids are 80°C and 20°C. When they are mixed in the mass ratio of 8:5, the temperature of the mixture becomes 60°C. Calculate the ratio of their specific heat capacities. (Neglecting heat lost to the surroundings)
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Given,
- Temperature of hot liquid = 80 °C
- Temperature of cold liquid = 20 °C
- Temperature of mixture = 60 °C
- Ratio of masses m1 : m2 = 8 : 5
Let the masses be m1 = 8m and m2 = 5m, and their specific heat capacities be c1 and c2 respectively.
Heat energy lost by the hot liquid = m1 x c1 x (80 - 60) = 8m x c1 x 20 = 160 m c1
Heat energy gained by the cold liquid = m2 x c2 x (60 - 20) = 5m x c2 x 40 = 200 m c2
By the principle of calorimetry,
Heat energy lost by the hot liquid = Heat energy gained by the cold liquid
$$\begin{aligned}160\ \text m\ \text c_1 = 200\ \text m\ \text c_2 \\[1em] \Rightarrow \frac{\text c_1}{\text c_2} = \frac{200}{160} = \frac{5}{4}\end{aligned}$$
Hence, the ratio of their specific heat capacities is 5 : 4.