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# Steam at 100oc is passed into 1.1 g of water contained in a calorimeter of water equivalent 0.02 g at 15oC till the temperature of the calorimeter and its contents rises to 80oC. The mass of the steam condensed in g is

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By Expert Faculty of Sri Chaitanya
a
0.130
b
0.065
c
0.260
d
0.135
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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 40oC. When m gram of ice at -10oC is put into the calorimeter and the mixture is allowed to attain equilibrium, the final temperature was found to be 20oC. It is known that specific heat capacity of the liquid changes with temperature as where $\mathrm{\theta }$ is temperature in oC. The specific heat capacity of ice, water and the calorimeter remains constant and values are  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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