Question

In an experiment of measuring specific heat of a liquid, a stream of liquid flows at a steady rate of $5g/s$ over an electrical heater dissipating $135W$ and a temperature rise of $5K$ is observed. On increasing the rate of flow to $10g/s$, the same temperature rise is produced with a dissipation of $235W$. Find the specific heat $\left(J/gK\right)$ of the liquid. (Assume heat loss to the surrounding in both the cases is same.)

Answer 1

We know that: $dQ=ms\Delta \theta$

Given:

$\text{Rate of heat loss}=135W$,
$\Delta \theta =5K$,
$\frac{dm}{d{t}_2}=10g/s$,
$\text{rate of heat loss}=235W$

Let $Q$ be heat absorbed per unit time which will be constant in both the cases
From $1^{st}$ case
$5\times s\times 5=135-Q$ ....(i)
From $2^{nd}$ case:
$10\times s\times 5=235–Q$ ....(ii)

Solving equations (i) and (ii), we get
$25s-135=50s-235$
$s=4J/gK$

Final Answer: 4