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 $0.8 .$ If the specific heat capacity of water is $1\, cal / gm$, the specific heat capacity of alcohol will be

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

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 $=V \times 1=V\, g$
Mass of alcohol $=V \times 0.8=0.8\, Vg$
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
$=\frac{1}{74}[V(50-40)+S \times 0.8 V \times 10]$
$=\frac{1}{74}[10 V+8 V \times S]$
Now $\frac{1}{74}[10 V+8 V \times S]=\frac{V}{5}$
$\therefore s=0.6\, cal / g -{ }^{\circ} C$