Question

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

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

Answer 1

Answer: (A)

Let $V$ be the volume of either liquid.
Thermal capacity of each calorimeter is $V$ cap per ${}^{\circ }C$
Mass of water $=V\times 1=Vg$
Mass of alcohol $=V\times 0.8=0.8Vg$
Rate of cooling of water,
$=\frac{1}{100}[V(50-40)+V\times 1\times (50-40)]=\frac{V}{5}cal/\sec$
Rate of cooling of alcohol
$=\frac{1}{74}[V(50-40)+S\times 0.8V\times 10]$
$=\frac{1}{74}[10V+8V\times S]$
Now $\frac{1}{74}[10V+8V\times S]=\frac{V}{5}$
$\therefore s=0.6cal/g-{}^{\circ }C$