# Science

• The temperature of a brass cylinder of mass 100g was raised to 100 degree Celsius and transferred to a thin aluminum can of negligible heat capacity containing 150g of paraffin at 11 degrees Celsius. If the final steady temperature after stirring was 20 degrees Celsius, calculate the specific heat capacity of paraffin.[Neglect heat losses, and assume specific heat capacity of brass =380J/g/K]

It appears that there's a typo in the problem. You might want to double check the source of the question and see if you have the correct information. The specific heat capacity of brass is 0.380 J/g-K. Either a decimal point is missing or the units are incorrect. 380 J/kg-K would be correct.

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The amount of heat lost by brass cylinder is transferred to paraffin. The amount of heat loss by the brass cylinder is given as

= mass x heat capacity x change in temperature

= 100 x 380 x (100 - 20) joules

The amount of heat gained by paraffin is given as

= mass x heat capacity x change in temperature

= 150 x heat capacity x (20 - 11)

Since amount of heat is constant,

100 x 380 x (100 - 20) = 150 x heat capacity x (20 - 11)

Solving this, we get the heat capacity of paraffin as 2251.8 J/kg/K.

In this solution, heat loss to aluminum is neglected since heat capacity of the aluminum can is negligible. Any other heat loss has also been neglected.

Hope this helps.

Approved by eNotes Editorial Team