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A piece of platinum of mass 200 g, is he...

A piece of platinum of mass 200 g, is heated for a sufficiently long time in a furnace and dropped quickly in 650 g of water at `10^(@)C`, kept in a container of water equivalent 50g. Final temperature of the mixture becomes `25^(@)C`. If the specific heat capacity of platinum is `"0.03 cal.g"^(-1).^(@)C^(-1)`, find the temperature of the furnace.

Text Solution

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Let the temperature of the furnace be `t^(@)C` which is alos that initial temperature of the piece of platinum.
Heat lost by the platinum piece
`=200xx0.03xx(t-25)=6(t-25)cal`
Heat gained by the container and water
`=50xx(25-10)+650xx1xx(25-10)`
`=750+9750="10500 cal"`
Assuming no loss of heat of the surroundings.
heat lost = heat gained
`"or, "6(t-25)=10500" or, "t=1775^(@)C`.
`therefore" Temperature of the furnace is "1775^(@)C`.
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