Aluminium container of mass of `10` g contains `200` g of ice at`-20^(@)C`. Heat is added to the system at the rate of `100` calories per second. What is the temperature of the system after four minutes? Draw a rough sketch showing the variation of the temperature of the system as a function of time . Given:
Specific heat of ice = `0.5"cal"g^(-1)(.^(@)C)^(-1)`
Specific heat of aluminium = `0.2"cal"g^(-1)(.^(@)C)^(-1)`
Latent heat of fusion of ice = `80"cal"g^(-1)`

Answer

`(##ALN_PHY_C07(I)_E01_253_A01##)`

Explanation

Heat needed to bring ice to freezing point
=`10 xx 0.2 xx 20 + 200 xx 0.5 xx 20`
=`40+2000 = 2040 "cal"`
Time taken to reach `0^(@)C = (2040)/(100) s = 20.4 s`
Heat needed to melt ice = `2040 + 200 xx 80 = 18040 "cal"`
Time taken to melt ice = `(18040)/(100) = 180.4 "sec"`
Heat taken in `4` min = `100 xx 4 xx 60 = 24000"cal"`
Let `theta` = final temperature then heat lost = heat gained `18040+10 xx 0.2 xx theta +200 xx1 xx (theta - 0) = 24000`
`implies theta = (24000-18040)/(202) = 29.50 ^(@)C`