Questions

Steam at 100^{o}c is passed into 1.1 g of water contained in a calorimeter of water equivalent 0.02 g at 15^{o}C till the temperature of the calorimeter and its contents rises to 80^{o}C. The mass of the steam condensed in g is

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a

0.130

b

0.065

c

0.260

d

0.135

NEW

detailed solution

Correct option is A

Heat is lost by steam in two stages: (i) for change of state from steam at 100oC to water at 100oC is m x 540(ii) to change water at 100oC to water at 80oC is m × 1 × (100 - 80), where m is the mass of steam condensed.Total heat lost by steam is m × 540 + m × 20 = 560m (cals) Heatgained by calorimeter and its contents is=(1.1+0.02)×(80−15)=1.12×65 cals using Principle of calorimetery, Heat gained = heat lost∴560m=1.12×65 ⇒m=0.130gm

Similar Questions

A calorimeter of water equivalent 10 g contains a liquid of mass 50 g at 40^{o}C. When m gram of ice at -10^{o}C is put into the calorimeter and the mixture is allowed to attain equilibrium, the final temperature was found to be 20^{o}C. It is known that specific heat capacity of the liquid changes with temperature as $\mathrm{S}=\left(1+\frac{\mathrm{\theta}}{500}\right)\text{cal}{{\mathrm{g}}^{-1}}^{\circ}{\mathrm{C}}^{-1}$where $\mathrm{\theta}$ is temperature in ^{o}C. The specific heat capacity of ice, water and the calorimeter remains constant and values are ${\mathrm{S}}_{\text{ice}}=0.5\text{cal}{{\mathrm{g}}^{-1}}^{\circ}{\mathrm{C}}^{-1}$; ${\mathrm{S}}_{\text{water}}=1.0\text{cal}{{\mathrm{g}}^{-1}}^{\circ}{\mathrm{C}}^{-1}$ and latent heat of fusion of ice is ${\mathrm{L}}_{\mathrm{f}}=80{\mathrm{calg}}^{-1}$. Assume no heat loss to the surrounding and calculate the value of m in grams.

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