1.

How much heat energy is necessary to raise the temperature of 2.5 kg of water from 20 ^@ C "to " 100^@C ?

Answer»

Solution :Given Mass (m) `=2.5 kg,` secific heat of water ( c)= 1 kcal /`kg^@C`
CHANGE in temperature `(Delta T)`
`=100 -20 = 80 ^@ C`
To FIND, heat ENERGY (Q)
Formula : `Q-m c Delta T`
Calculation: According to priniciple of heat exchange , Energy supplied to water
= Energy gained by water
From formula,
`Q=2.5 xx 1 xx 80 `
= 400 kcal.


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