3yes i did it ur way ...... seems dat ive made a cac.
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
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8 Answers
21GUYS limits that u hav applied in integration are WRONG!!!
TEMP IS IN KELVIN not CELSIUS!!!!!!
2Anyone here can give me the complete solution to this problem ????
11Heat gained by the ice = 0.1*8*104 + 27*1000
= 35*103 cal
Heat Lost by the container =
mSdT
dQ =mSdt
dQ =27∫227 m*(A+BT)dt
dQ =27∫227madt +27∫227BTdt
Q = m*A*(200) + B(200)(254)/2
Q = Heat gained by the ice
35 * 1000 = 200(m * 100 + 0.01 *254)
35*5 = 100m +2.54
100m = 172.46
m = 1.7246kg
Significant digit is 4 Therefore mass of the container is 1.724kg