Question

$250g$ of water and equal volume of alcohol of mass $200g$ are replaced successively in the same calorimeter and cool from ${60}^{\circ }$ to ${55}^{\circ }$ in $130s$ and $67s$ respectively. If the water equivalent of the calorimeter is $10g$, then the specific heat of alcohol in $cal/g^{\circ }C$ is.

• A. $1.25$
• B. $0.69$
• C. $0.62$
• D. $0.68$

Answer 1

The correct option is D $0.68$

Firstly, we have to calculate the temperature change per second during the process in the calorimetry.

For water,

$\Delta t_w=\frac{{\theta }_1-{\theta }_2}{t_w}$

where ${\theta }_1=$ initial temperature

${\theta }_2=$ final temperature

$t_w=$ time taken by water $=\frac{60-55}{130}={\frac{5}{130}}^{o}C/sr$

For alcohol,

$\Delta t_a=\frac{{\theta }_1-{\theta }_2}{t_a}$

where $t_a=$ time taken by alcohol

So, $\Delta t_a=\frac{60-55}{67}={\frac{5}{67}}^{o}C/sr$

Now, by applying principle of calorimetry,

Heat gained $=$ Heat lost

So, $(m+m_1)s_1\Delta t_w=m_2{s}_2\Delta t_a$

where $m=$ Mass of calorimeter, $m_1=$ Mass of water, $m_2=$ Mass of alcohol, $s_1=$ Specific heat of water, $s_2=$ Specific heat of alcohol.

Now, $m=10g,m_1=250g,m_2=200g,s_1=1cal/g{/}^{o}C$

So, $(250+10)\times 1\times \frac{5}{130}=200\times s_2\times \frac{5}{67}$

So, $s_2=0.68cal/g{/}^{o}C$