JEE Main20203 Sep 2020Evening ShiftPhysicsThermal Properties of MatterActual
A calorimeter of water equivalent 20 g contains 180 g of water at 25 ° C . m ' ' ' grams of steam at 100 ° C is mixed in it till the temperature of the mixture is 31 ° C . The value of m ' ' is close to (Latent heat of water = 540 cal g - 1 , specific heat of water = 1 cal g - 1 ° C - 1 )
Options
- A2
- B4
- C3 . 2
- D2 . 6
Correct answer
A. 2
Step-by-step solution
200 31 - 25 = m × 540 + m 1 69 1200 = m 609 m ≈ 2