Question

A liquid having mass $m=250\text{g}$ is kept warm in a vessel by using an electric heater. The liquid is maintained at ${50}^{\circ }\text{C}$ when the power supplied by the heater is $30\text{W}$ and the surrounding temperature is ${20}^{\circ }\text{C}$. As the heater is switched off, the liquid starts cooling and it was observed that it takes $10\text{sec}$ for temperature to fall from ${40}^{\circ }\text{C}$ to ${39.9}^{\circ }\text{C}$. Calculate the specific heat capacity of the liquid. Assume Newton's law of cooling to be applicable.

• A. $6500{\text{J/kg}}^{\circ }\text{C}$
• B. $7473{\text{J/kg}}^{\circ }\text{C}$
• C. $7980{\text{J/kg}}^{\circ }\text{C}$
• D. $8493{\text{J/kg}}^{\circ }\text{C}$

Answer 1

The correct option is C $7980{\text{J/kg}}^{\circ }\text{C}$

From Newton's law of cooling, we know that
$\frac{dQ}{dt}=k(\theta -{\theta }_0)$, where ${}^{\prime }{s}^{\prime }$ is specific heat, ${}^{\prime }{\theta }^{\prime }$ is mean temperature, ${}^{\prime }{\theta }_0^{\prime }$ is surrounding temperature and $k$ is a constant.
From the question, we can say that power supplied by heater is equal to the rate of heat loss from the liquid to the surroundings [so as to maintain constant temperature]
i.e $30\text{W}=k(50-20)$
$\Rightarrow k=1\text{W}{/}^{\circ }\text{C}....\left(1\right)$

After the heater is switched off, rate of loss of heat of liquid
$\frac{dQ}{dt}=ms\frac{d\theta }{dt}=k(\theta -{\theta }_0)$
For fall in temperature from ${40}^{\circ }\text{C}$ to ${39.9}^{\circ }\text{C}$ in $10\text{s}$,
Mean temperature $={39.95}^{\circ }\text{C}$
$ms\frac{0.1}{10}=k(39.95-20)$
Using$\left(1\right)$, we get
$0.25\times s\times 0.01=1\times 19.95$
$\Rightarrow s=7980{\text{J/kg}}^{\circ }\text{C}$
Thus, option (c) is the correct answer.