Verified Solution • 2025–26 Session
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QJEEM20Phb21c: JEE Main 2020 Physics - Class 11 JEEM20Phb21c Solution: A calorimeter of water equivalent 20 g contains 180 g of water…
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)
Multiple Choice Options
- A 2.6
- B 2
- C 4
- D 3.2
Step-by-Step Verified Solution
Given Temp of mixture = 31°C
180×1×(31-25) + 20×(31-25) = m×540 + m×1×(100-31)
180×6 + 20×6 = 540m + 100 m - 31m
1080 + 120 = 640 m - 31m
1200 = 609m
m = = 1.97
180×1×(31-25) + 20×(31-25) = m×540 + m×1×(100-31)
180×6 + 20×6 = 540m + 100 m - 31m
1080 + 120 = 640 m - 31m
1200 = 609m
m = = 1.97