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A resistance heater was wound around a 5...

A resistance heater was wound around a `5.0g` metallic cylinder. A current of `0.84 A` was passed through the heater for `20s` while the drop in voltage across the heater was `50V`. The temperature change of the cylinder was from `25^(@)C` before the heating period and `35^(@)C` at the end. If the heat loss is neglected, what is the specific heat of the cylinder metal in `cal g^(-1)K^(-1)`.

Text Solution

Verified by Experts

Energy input `=I.V.t`
`=(0.84A)(5V)(20s)`
`=(84J)/(4.18J cal^(-1))~~20calo rie`
Also
Energy input `=` Mass `xx` Specific heat `xx` Temperature rise ltbr. `=(5.0g)(` Specific heat `)xx[(35-25)K]`
`=50xx` specific heat
`20 cal =50 xx `specific heat
Specific heat `=(20)/(50)=0.4 cal g^(-1) K^(-1)`
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