Question

Equal volumes of water and alcohol when put in similar calorimeters take $100\text{sec}$ and $74\text{sec}$ respectively, to cool from ${50}^{\circ }\text{C}$ to ${40}^{\circ }\text{C}$. The thermal capacity of each calorimeter is numerically equal to the volume of either liquid. The specific gravity of alcohol is 0.8. If the specific heat capacity of water is $1\text{cal/g}{}^{\circ }\text{C}$, the specific heat capacity of alcohol will be -
[Assume constant rate of cooling]

• A. $0.6\text{cal/g}{}^{\circ }\text{C}$
• B. $0.8\text{cal/g}{}^{\circ }\text{C}$
• C. $1.6\text{cal/g}{}^{\circ }\text{C}$
• D. $1.8\text{cal/g}{}^{\circ }\text{C}$

Answer 1

The correct option is A $0.6\text{cal/g}{}^{\circ }\text{C}$

Let $V\text{(in cc)}$ be the volume of either liquid.
Then, thermal capacity of each calorimeter is $V\text{cal}{/}^{\circ }\text{C}$
Given,
Mass of water $m_w=V\times {\rho }_w=V\times 1=V\text{grams}$
Mass of alcohol $m_a=V\times {\rho }_a=V\times 0.8=0.8V\text{grams}$
Time taken by water to cool from ${50}^{\circ }\text{C}$ to ${40}^{\circ }\text{C}$ is $\left(\Delta t_1\right)=100\text{sec}$
Time taken by alcohol to cool from ${50}^{\circ }\text{C}$ to ${40}^{\circ }\text{C}$ is $\left(\Delta t_2\right)=74\text{sec}$
We know that,
Rate of cooling of water $\frac{(-d{Q}_w)}{dt}=\frac{\text{Heat released by water}}{\text{Time taken}}+\frac{\text{Heat released by calorimeter-1}}{\text{Time taken}}$
$=(m_w{c}_w\times \frac{\Delta T_w}{\Delta t_1})+(m_v{c}_v\times \frac{\Delta T_v}{\Delta t_1})$
$\Rightarrow \frac{(-d{Q}_w)}{dt}=\frac{1}{100}[V\times 1\times (50-40)+V\times (50-40\left)\right]$
[given $m_w=V,m_v{c}_v=V$]
$=\frac{V}{5}\text{cal/sec}$

Rate of cooling of alcohol $\frac{(-d{Q}_a)}{dt}=\frac{\text{Heat released by alcohol}}{\text{Time taken}}+\frac{\text{Heat released by calorimeter-2}}{\text{Time taken}}$
$=(m_a{c}_a\times \frac{\Delta T_a}{\Delta t_2})+(m_v{c}_v\times \frac{\Delta T_v}{\Delta t_2})$
$\Rightarrow \frac{(-d{Q}_a)}{dt}=\frac{1}{74}[s\times 0.8V\times (50-40)+V\times (50-40\left)\right]$
[given $m_a=0.8V,m_v{c}_v=V,s=$specific heat capacity of alcohol]
$=\frac{1}{74}[10V+8V\times s]\text{cal/sec}$

Given, Rate of cooling of alcohol = Rate of cooling of water
$\Rightarrow \frac{1}{74}[10V+8V\times s]=\frac{V}{5}$
$\therefore s=0.6\text{cal/g}{}^{\circ }\text{C}$
Thus, option (a) is the correct answer.