Question

A flow calorimeteris a device used to measure the specific heat of a liquid. Energy is added as heat at a known rate to a stream of the liquid as it passes through the calorimeter at a known rate. Measurement of the resulting temperature difference between the inflow and the outflow points of the liquid stream enables us to compute the specific heat of the liquid. Suppose a liquid of density$\mathbf{0}\mathbf{.85}\mathbf{}{\mathbf{\text{g/cm}}}^{\mathbf{\text{3}}}$ flows through a calorimeter at the rate of$\mathbf{8}\mathbf{.0}{\mathbf{\text{\hspace{0.17em}cm}}}^{\mathbf{\text{3}}}\mathbf{\text{/s}}$ .When energy is added at the rate of$\mathbf{250}\mathbf{}\mathbf{\text{W}}\mathbf{}$ by means of an electric heating coil, a temperature difference of $\mathbf{15}{\mathbf{}}^{\mathbf{\text{0}}}\mathbf{\text{C}}$is established in steady-state conditions between the inflow and the outflow points.What is the specific heat of the liquid?

Short answer

The specific heat of the liquid is $2.5\times {10}^3\frac{\text{J}}{\text{kg°C}}$.

Step-by-step solution

Stating thegiven data

1. The density of the liquid is$\rho =0.85\frac{\text{g}}{{\text{cm}}^{\text{3}}}$ .
2. The volume of the liquid flowing through calorimeter per second is$V=8.0\frac{{\text{cm}}^{\text{3}}}{\text{s}}$.
3. The rate of adding or takeaway of energy from the calorimeter is$P=250\text{W}$.
4. The temperature difference between the inflow and outflow is $\Delta T=15\text{°C}$.

Understanding the concept of specific heat

The amount of heat required to increase the temperature of 1 gram of a substance by 1 degree Celsius is known as its specific heat. We can use the concept of specific heat of the water. We can use the expression of density and find the mass of the liquid. By using the concept of power, we can find the amount of heat that this mass takes away from the calorimeter.

Formulae:

Density of the body,$\rho =\frac{m}{V}$…(i)

Rate of radiation by the body via thermal radiation,$P=\frac{Q}{t}$…(ii)

Heat transferred by the body via thermal energy,$Q=cm(T_f-T_i)$ .…(iii)

Calculation of the specific heat of the liquid

The system is in a steady state; hence, the amount of heat added to the liquid, in one second, from the heating coil is equal to the amount of heat flowing away from the liquid in one second through the calorimeter.

The volume of the liquid flowing through the calorimeter in one second is$V=8.0{\text{cm}}^{\text{3}}$.

From equation (i), the value of the mass of the body is given by

$\begin{array}{c}m=0.85\frac{\text{g}}{{\text{cm}}^{\text{3}}}\times 8.0{\text{cm}}^{\text{3}}\\ =6.8\text{g}\end{array}$

In one second, the amount of heat that this mass takes away from the calorimeter using equation (ii) is given by

$\begin{array}{c}Q=250\text{W}\times 1\text{s}\\ =250\text{J}\end{array}$

The expression for the specific heat of the liquid using equation (iii) is given by

$\begin{array}{c}c=\frac{Q}{m\Delta T}\\ =\frac{250\text{J}}{6.8\text{g}\times 15\text{°C}}\\ =2.45\frac{\text{J}}{\text{g°C}}\\ \approx 2.5\times {10}^3\frac{\text{J}}{\text{kg°C}}\end{array}$

Hence, the value of the specific heat of the liquid is $2.5\times {10}^3\frac{\text{J}}{\text{kg°C}}$.