How much heat energy is necessary to raise the temperature of 5 kg of water from `20 ^@ C " to " 100^@C ? `

Explanation

Given Mass (m) `=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=5 xx 1 xx 80 `
= 400 kcal.