Question

A container of negligible heat capacity contains $1\text{kg}$ water. It is connected by a steel rod of length $1\text{m}$ and area of cross section $1{\text{m}}^2$ to a large steam chamber which is maintained at ${100}^{\circ }\text{C}$. If the initial temperature of water is $0^{\circ }\text{C}$, find the time (in seconds) after which it reaches ${50}^{\circ }\text{C}$ .
[Neglect heat capacity of steel rod and assume no heat is lost to surroundings. Also, take specific heat of water $=S{\text{J/kg}}^{\circ }\text{C}$, conductivity of steel $k_{steel}=k{\text{W/m}}^{\circ }\text{C}]$

• A. $\frac{S}{k}\ln 2$
• B. $\frac{S}{2k}\ln 2$
• C. $\frac{2S}{k}\ln 2$
• D. $\frac{2k}{S}\ln 2$

Answer 1

The correct option is A $\frac{S}{k}\ln 2$

Let the temperature of water at time $t$ be $T$.
Thermal current, $I=\frac{(100-T)}{R}$
where $R$ = thermal resistance.
$R=\frac{L}{kA}......\left(1\right)$
For increase in temperature of water from $T\to T+dT$
$I=\frac{dQ}{dt}=mS\frac{dT}{dt}$
$\Rightarrow \frac{100-T}{R}=mS\frac{dT}{dt}$
Integrating on both sides, we get
$\Rightarrow {\int }_0^{50}\frac{dT}{100-T}={\int }_0^t\frac{dt}{mSR}$
$\Rightarrow -\ln (100-50)+\ln (100-0)=\frac{t}{mSR}$
$\Rightarrow \ln 2=\frac{t}{mSR}$
$\Rightarrow t=mSR\ln 2$

From $\left(1\right)$, we get,
$t=\frac{L}{kA}mS\ln 2$
Given, $m=1\text{kg}$; $L=1\text{m}$ and $A=1{\text{m}}^2$
We can write the above equation as,
$t=\frac{S}{k}\text{ln}2\text{sec}$
Thus, option (a) is the correct answer.