Question

Two identical containers of the same emissivity containing liquids A & B at the same temperature of $60°C$ initially and densities ${\rho }_A$ and ${\rho }_B$ respectively. Where, ${\rho }_A<{\rho }_B$. Which plot best represents the temperature variation of both with time? Given

$$\begin{gathered} S_A=100\frac{J}{Kg-K} \\ {S}_B=2000\frac{J}{Kg-K} \end{gathered}$$

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

Step 1: Given data

Density of container A = ${\rho }_A$

Density of container B = ${\rho }_B$

Temperature of both containers = $60°C$

Both containers have the same emissivity also ${\rho }_A<{\rho }_B$.

Step 2: Calculation

The rate of change of temperature is given by the expression,

$\frac{d\theta }{dt}=\frac{4\sigma eA{{\theta }_0}^3(\theta -{\theta }_0)}{ms}$ ……………………. equation 1

where $m=mass$, ${\theta }_0=Temperatureofbody$, $\theta =Temperatureofsurroundings$ and $\sigma ,A,e$ are constants.

The above expression for $t=0$ can be written as $\frac{d\theta }{dt}=\frac{1}{ms}$, so the graph will be inversely proportional to mass.

Since $S_A<S_B$ and $m_A<m_B$ at $t=0$, the graph of A will be below B.

Here $S_A=$ surrounding temperature of A

$S_B=$ surrounding temperature of B

Therefore, the graph obtained in option (B) is the correct option.