# IIT JEE past question Thermodynamics

An Ice cube of mass 0.1 kg at 0 degree C is placed in an isolated container which is at 227 degree C. the specific heat s o fthe container varies with temperature T according to the empirical relation S=A+BT. where A=100cal/kg-K and B=2x10-2 cal/Kg-K2.
If the final temperature of the container is 27degree C, determine the mass of the container.

Latent heat of fusion for water =8x104 cal/Kg-K, Specific heat of water is 1000 cal/Kg-K

21
tapanmast Vora ·

thanx Sir!!! [1] Thermo is EHRE!!!

IS da ans = 0.495kg?

62
Lokesh Verma ·

3
iitimcomin ·

yes i did it ur way ...... seems dat ive made a cac.

21
tapanmast Vora ·

GUYS limits that u hav applied in integration are WRONG!!!

TEMP IS IN KELVIN not CELSIUS!!!!!!

11
virang1 Jhaveri ·

ok thx for pointing out my mistake

3
iitimcomin ·

i did kelv only ..... i have bad multi. skills....

2
Shaurya Dixit ·

Anyone here can give me the complete solution to this problem ????

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virang1 Jhaveri ·

Heat gained by the ice = 0.1*8*104 + 27*1000
= 35*103 cal
Heat Lost by the container =
mSdT
dQ =mSdt
dQ =27âˆ«227 m*(A+BT)dt