Question

A metal vessel of negligible heat capacity contains $0.5\text{kg}$ of water (of specific heat capacity $4200{\text{J kg}}^{-1}{\text{K}}^{-1})$. It is heated from ${15}^{\circ }\text{C}$ to ${85}^{\circ }\text{C}$ by a $750\text{W}$ immersion heater in 4 minutes. The average loss of heat in watts to the surroundings, from the vessel during this time, is nearly:

• A. $70$
• B. $123$
• C. $138$
• D. $212$

Answer 1

The correct option is C $138$

Given,
Mass of water $\left(m\right)=0.5\text{kg}$
Specific heat capacity of water $\left(s\right)=4200{\text{J kg}}^{-1}{\text{K}}^{-1}$
Rise in temperature $\left(\Delta \theta \right)={70}^{\circ }\text{C}$

Power of heater $P=750\text{W}$
$P=\frac{W}{t}\Rightarrow W=Pt....\left(1\right)$
(where $W=$ work done by heater)

Time of heating = 4 minutes $=4\times 60\text{sec}$
Average loss of heat $\left(Q_{\text{loss}}\right)=?$

Work done by the heater will be equal to the total heat absorbed by system + heat dissipated
$Pt=Q_{\text{system}}+Q_{loss}$
$\Rightarrow Pt=ms\Delta \theta +Q_{loss}$
$\Rightarrow 750\times 4\times 60=0.5\times 4200\times 70+Q_{\text{loss}}$
$\Rightarrow Q_{\text{loss}}=3.3\times {10}^4$

Now, the rate of heat loss in 4 minutes
$P_{\text{loss}}=\frac{Q_{loss}}{t}=\frac{3.3\times {10}^4}{4\times 60}$
$=137.50\approx 138\text{W}$