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 $1 cal/g^{\circ} C$ , the specific heat capacity of alcohol will be -
[Assume constant rate of cooling]

0.6 cal/g^{\circ} C

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0.8 cal/g^{\circ} C

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1.6 cal/g^{\circ} C

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1.8 cal/g^{\circ} C

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Solution

The correct option is A $0.6 cal/g^{\circ} C$
Let $V (in cc)$ be the volume of either liquid.
Then, thermal capacity of each calorimeter is $V cal /^{\circ} C$
Given,
Mass of water $m_{w} = V \times ρ_{w} = V \times 1 = V grams$
Mass of alcohol $m_{a} = V \times ρ_{a} = V \times 0.8 = 0.8 V grams$
Time taken by water to cool from $50^{\circ} C$ to $40^{\circ} C$ is $(Δ t_{1}) = 100 sec$
Time taken by alcohol to cool from $50^{\circ} C$ to $40^{\circ} C$ is $(Δ t_{2}) = 74 sec$
We know that,
Rate of cooling of water $\frac{(- d Q_{w})}{d t} = \frac{Heat released by water}{Time taken} + \frac{Heat released by calorimeter-1}{Time taken}$
$= (m_{w} c_{w} \times \frac{Δ T_{w}}{Δ t_{1}}) + (m_{v} c_{v} \times \frac{Δ T_{v}}{Δ t_{1}})$
$\Rightarrow \frac{(- d Q_{w})}{d t} = \frac{1}{100} [V \times 1 \times (50 - 40) + V \times (50 - 40)]$
[given $m_{w} = V, m_{v} c_{v} = V$ ]
$= \frac{V}{5} cal/sec$

Rate of cooling of alcohol $\frac{(- d Q_{a})}{d t} = \frac{Heat released by alcohol}{Time taken} + \frac{Heat released by calorimeter-2}{Time taken}$
$= (m_{a} c_{a} \times \frac{Δ T_{a}}{Δ t_{2}}) + (m_{v} c_{v} \times \frac{Δ T_{v}}{Δ t_{2}})$
$\Rightarrow \frac{(- d Q_{a})}{d t} = \frac{1}{74} [s \times 0.8 V \times (50 - 40) + V \times (50 - 40)]$
[given $m_{a} = 0.8 V, m_{v} c_{v} = V, s =$ specific heat capacity of alcohol]
$= \frac{1}{74} [10 V + 8 V \times s] cal/sec$

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

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Similar questions

Q. Equal volumes of water and alcohol when put in similar calorimeters take

100 sec

and

74 sec

respectively, to cool from

50^{\circ} C

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/g^{\circ} C

, the specific heat capacity of alcohol will be -
[Assume constant rate of cooling]

Q. 250 g of water and equal volume of alcohol of mass 200 g are replaced successively in the same calorimeter and cools from

60^{\circ} C

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in 130 s and 67 s, respectively. If the water equivalent of the calorimeter is 10 g and rate of heat loss is same for both, then the specific heat of alcohol in

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are replaced successively in the same calorimeter and cool from

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130 s

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respectively. If the water equivalent of the calorimeter is

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, then the specific heat of alcohol in

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