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A vacuum flask contain 0.3 kg of liquid ...

A vacuum flask contain 0.3 kg of liquid paraffin whose temperature can be increased at the rate of `1^(@)C` per minute using an immersion heater of power 12.3W. When a 19.2 W heater is used to heat 0.4 kg of liquid paraffin, in same flask, the rate of rise of temperature is `1.2^(@)C` per minute. Find the specific heat of paraffin and the thermal capacity of the flask.

Text Solution

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Let s = specific heat of paraffin,C = thermal capacity of the flask.
Heat developed per minute by the 1st heater
`=(12.3xx60)/(4.2)"cal = 175.71 cal"`
`" [4.2J of energy = 1 cal of heat]"`
Similarly, heat developed per minute by the second heater
`=(19.2xx60)/(4.2)=274.28cal`
From the given data,
heat gained by paraffin in the first case = heat supplied by the heater
`therefore" "03xx1000xxsxx1+Cxx1=175.71`
`"or, "300s+C=175.71" ...(1)"`
and similarly, for the heat gained by paraffin and the flask in the second case,
`0.4xx1000xxsxx1+Cxx1=274.28`
`"or, "480s+1.2C=274.28" ...(2)"`
Solving (1) and (2), we get
`s="0.529 cal.g"^(-1).^(@)C^(-1)`
`"and "C="17 cal".^(@)C^(-1)" (approx)"`
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