Question

A liquid of density $0.85 \, g \, cm^{- 3}$ flows through a calorimeter at the rate of $8.0 \, cm^{3} \, s^{- 1}$ . Heat is added by means of a $250 \, W$ 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 ( $1 \, kcal \, =4186 \, J$ )

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

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 = $Q_{s}$ $=$ $250J$ $= \frac{2 5 0}{4 1 8 6} \text{kcal}$
The volume of liquid flowing out per second = $8.0cm^{3}=8\times 10^{- 6}m^{3}$
Mass of this liquid = $\left(\right.0.85\left.\right)\times 1000\times 8\times \left(10\right)^{- 6}kg$
The temperature rise of this mass of liquid = $15^\circ C$
Hence, $= \frac{2 5 0}{4 1 8 6} = \text{mst} = \text{0.85} \times 8 \times 1 0^{- 3} \times \text{s} \times 1 5$
Hence, $\text{s}=\frac{2 5 0 \times 1 0^{3}}{4 1 8 6 \times \text{0.85} \times 8 \times 1 5}=\text{0.6 }kcalkg^{- 1}K^{- 1}$