Question

A liquid of density $0.85gc{m}^{-3}$ flows through a calorimeter at the rate of $8.0c{m}^3{s}^{-1}$ . Heat is added by means of a $250W$ electric heating coil and a temperature difference of $15^{\circ }C$ is established in steady-state conditions between the inflow and the outflow points of the liquid. The specific heat for the liquid will be ( $1kcal=4186J$ )

• A. $0.6kcalk{g}^{-1}{K}^{-1}$
• B. $0.3kcalk{g}^{-1}{K}^{-1}$
• C. $0.5kcalk{g}^{-1}{K}^{-1}$
• D. $0.4kcalk{g}^{-1}{K}^{-1}$

Answer 1

Answer: (A)

This is a problem on 'flow calorimeter' used to measure the specific heat of the liquid.
Amount of heat supplied to the water per second by the heating coil = $Q_s$ $=$ $250J$ $=\frac{250}{4186}\text{kcal}$
The volume of liquid flowing out per second = $8.0c{m}^3=8\times 10^{-6}{m}^3$
Mass of this liquid = $(0.85)\times 1000\times 8\times {\left(10\right)}^{-6}kg$
The temperature rise of this mass of liquid = $15^{\circ }C$
Hence, $=\frac{250}{4186}=\text{mst}=\text{0.85}\times 8\times 10^{-3}\times \text{s}\times 15$
Hence, $\text{s}=\frac{250\times 10^3}{4186\times \text{0.85}\times 8\times 15}=\text{0.6 }kcalk{g}^{-1}{K}^{-1}$