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Thermal Physics

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25 Dec 2007 17:43:53 IST

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calorimetry question

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A mixture of 250gm of water and 200gm of ice is kept in a calorimeter of water equivalent 50gm. If 200gm of steam at 100^oC is passed through this mixture. Calculate the final temperature and the weight of the container of the calorimeter.

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ramyani chakrabarty

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Joined: 22 Apr 2007

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26 Dec 2007 15:29:14 IST

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heat given out by 200g of steam to be converted to water = 108000 cal

now heat reqd for the water present in the calorimeter to reach the temp of 100 deg from 0 degree (since it is a water ice mixture, its temp is at 0 degree) = 250 x 1x 100 = 25000 cal
heat reqd for the calorimeter to reach the temp of 100 deg= 50 x 100= 5000 cal

heat reqd for the ice to melt = 200 x 80 = 16000 cal

heat reqd by the melted ice ( which is the water) to reach
the temperature of 100 deg = 200 x 1 x 100= 20000 cal


so total heat reqd by the calorimeter along with the water ice mixture to reach the temp of 100 deg         = 25000 + 5000 + 16000 + 20000 = 66000 cal, which is less than the heat supplied by the condensation of 200 g of steam. so only a part of the steam wil condense which will supply 66000 cal of heat reqd for the above processes.the final temperature of the mixture will be 100 degree celsius, and the mixture will be composed of water and steam at 100 degree celsius.

Bipin Dubey

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Joined: 23 Jan 2007

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2 Jan 2008 23:51:29 IST

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Good work Ramyani.

ramyani chakrabarty

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2 Jan 2008 23:59:00 IST

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Thank you sir !

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