JEE MAIN - Physics (2020 - 3rd September Evening Slot - No. 5)
A calorimeter of water equivalent 20 g contains 180 g of water at 25oC. ‘m’ grams of steam at
100oC is mixed in it till the temperature of the mixure is 31oC. The value of ‘m’ is close to :
(Latent heat of water = 540 cal g–1, specific heat of water = 1 cal g–1 oC–1)
(Latent heat of water = 540 cal g–1, specific heat of water = 1 cal g–1 oC–1)
2.6
2
4
3.2
Explanation
Given Temp of mixture = 31°C
180×1×(31-25) + 20×(31-25) = m×540 + m×1×(100-31)
$$ \Rightarrow $$ 180×6 + 20×6 = 540m + 100 m - 31m
$$ \Rightarrow $$ 1080 + 120 = 640 m - 31m
$$ \Rightarrow $$ 1200 = 609m
$$ \Rightarrow $$ m = $${{1200} \over {609}}$$ = 1.97
180×1×(31-25) + 20×(31-25) = m×540 + m×1×(100-31)
$$ \Rightarrow $$ 180×6 + 20×6 = 540m + 100 m - 31m
$$ \Rightarrow $$ 1080 + 120 = 640 m - 31m
$$ \Rightarrow $$ 1200 = 609m
$$ \Rightarrow $$ m = $${{1200} \over {609}}$$ = 1.97
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