## Related questions with answers

A flow calorimeter is 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 $0.85 \mathrm{~g} / \mathrm{cm}^3$ flows through a calorimeter at the rate of $8.0 \mathrm{~cm}^3 / \mathrm{s}$. When energy is added at the rate of $250 \mathrm{~W}$ by means of an electric heating coil, a temperature difference of $15 \mathrm{C}^{\circ}$ is established in steady-state conditions between the inflow and the outflow points. What is the specific heat of the liquid?

Solutions

VerifiedIt is given that the density of the liquid is $\rho = 0.85\ \frac{\text{g}}{\text{cm}^{3}}$, volume of the liquid through the calorimeter is $V= 8\ \text{cm}^{3}$, heat/energy added is $Q = 250$ Joule.

Since change in temperature when heat is added $\Delta T= 15^{\text{\textdegree}}$ C, to find the specific heat of the liquid $c$, we use the Energy Formula as

$\begin{align*} Q & = mc \Delta T\\ c & = \frac{Q}{m\Delta T}\\ & = \frac{Q}{\rho V \Delta T}\\ & = \frac{250}{0.85 \cdot 8 \cdot 15}\\ & = \boxed{2.45}\ \frac{\text{Joule}}{\text{Kg} \cdot \text{C}}\\ \end{align*}$

Therefore, the specific heat of the liquid is $\textbf{2.45}\ \frac{\text{Joule}}{\text{Kg} \cdot \text{C}}$

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