Question

$250\text{gm}$ of water and equal volume of alcohol of mass $200\text{gm}$ are replaced successively in the same calorimeter which cools from ${60}^{\circ }C$ to ${55}^{\circ }C$ in $130$ seconds and $67$ seconds respectively. If the water equivalent of the calorimeter is $10\text{gm}$, then the specific heat of alcohol (in ${\text{cal/g}}^{\circ }C$) is
(Assume that rate of heat loss in both the cases remains same)

• A. $1.30$
• B. $0.67$
• C. $0.62$
• D. $0.985$

Answer 1

The correct option is C $0.62$

Mass of alcohol $=200$ gm
When water is added to calorimeter,
Given water equivalent of calorimeter, $W=10$ gm
Fall of temperature $=60-55=5^{\circ }C$
Time taken by water to cool $=130$ sec.
Heat lost by water and calorimeter
$=(250+10)\left(1\right)5=260\times 5=1300$ cal
Rate of loss of heat $=10\frac{\text{cal}}{\text{sec}}$

When alcohol is added,
Heat lost by alcohol and calorimeter is $=(200s+10)5\text{cal}$
Heat lost by alcohol and calorimeter is $=\frac{(200s+10)5}{67}\frac{\text{cal}}{\text{sec}}$


Rates of loss of heat in the two cases are equal
$\therefore \frac{(200s+10)5}{67}=10$
$1000s+50=670$
$1000s=620$
$s=0.62\frac{\text{cal}}{g^{\circ }C}$
Hence, the correct answer is option (c).