Question

Two identical calorimeters 'X' and 'Y', of water equivalent 10 g each, contain equal quantities of the same liquid. The mass, initial temperature and specific heat capacity of the liquid present in both calorimeters are 50 g, ${25}^{\circ }C$ and $\frac{1}{2}cal{g}^{-1}{}^{\circ }{C}^{-1}$ respectively. A 10 g metal piece (say A) of specific heat capacity $0.4cal{g}^{-1}{}^{\circ }{C}^{-1}$ is dropped in calorimeter "X". and 25 g of metal piece (say B) of specific heat capacity "$S_B$" is dropped into calorimeter "Y". Due to this, the equilibrium temperature in 'X' and 'Y' rises to ${30}^{\circ }C$ and ${40}^{\circ }C$ respectively. If the initial temperature of 'B' is twice that of 'A', then find the value of "$S_B$".

Answer 1

X	Y
Water equivalent =10gm Mass of liquid in x = 50gm Temperature = 25°C Specific heat of liquid $=\frac{1}{2}cal/gm°C$	Water equivalent = 10gm Mass of liquid in y= 50gm Temperature = 25°C Specific heat capacity of liquid $=\frac{1}{2}cal/gm°C$

Mass of A metal substance dropped in calorimeter $X=10$gm

Specific heat capacity of metal $A=0.4$ cal/gm°C

Mass of B metal substance dropped in calorimeter $Y=25$gm

Specific heat capacity of metal $B=?$ $=S_B$

Given: initial temperature of B is twice the temperature of A

so let the temperature of $A$ be $T°C$

and the temperature of $B$ be $2T°C$

and Given: Final temperature of $X$ with $A=30°C$

Final temperature of $Y$ with $B=40°C$

As $X$ and $A$ are in thermal equilibrium

$Y$ and $B$ are in thermal equilibrium

So using the principle of calorimeter

Heat gained by calorimeter = Heat lost by metal

So for now $X$

Heat gained by calorimeter$X+$ liquid in $X$ = Heat lost by metal A $\left(1\right)$

Now firstly we get heat gained by calorimeter $X$

Heat gained by calorimeter is equivalent to heat given to water of $10$gm to increase the temperature from $25°C$ to $30°C$ $=C$

$Q_1=mc\Delta T$

$Q_1=10\times 1\times 5$

$Q_1=50calories$

Heat required to increase the temperature of liquid

$Q_2=mc\Delta T$

$Q_2=50\times \frac{1}{2}\times 5$

$Q_2=125calories$

Total heat gained by calorimeter + liquid $=Q_1+Q_2=175calories$

According to equation (1)

Heat lost by metal A =$175calories$

$mc\Delta T=175calories$

$10\times 0.4\times (T-30)=175$

$4T-120=175$

$4T=295$

$T={\left(\frac{295}{4}\right)}^{0}C$

so initial temperature of A metal $={\left(\frac{295}{4}\right)}^{0}C$

and initial temperature of B metal $={\left(\frac{295}{2}\right)}^{0}C$

Now for $Y$ calorimeter

Heat gained by calorimeter $Y$ $+$ liquid = Heat lost by metal B $\left(2\right)$

Now heat gained by calorimeter = heat given to water of 10gm to increase the temperature from$25°C$ to $45°C$ $=Q_3$

$=Q_3=mc\Delta T$

$=Q_3=10\times 1\times 15$

$=Q_3=150calories$

Heat gained by liquid in $Y$

$=Q_4=mc\Delta T$

$=50\times \frac{1}{2}\times 15$

$=Q_4=375calories$

total heat gained by calorimeter $Y$ + liquid $=Q_3+Q_4=525calories$

Using the equation (2)

Heat lost by metal B = $525$ calories

$mc\Delta T$ = $525$ calories

$25\times S_B\times (\frac{295}{2}-40)=525$

$25\times S_B\times \frac{215}{2}=525$

$S_B=\frac{525\times 2}{25\times 215}$

$S_B=\left(\frac{42}{215}\right)cal/gm°C$