Calculate the amount of heat energy required to increase the temperature of 250 g of water from `27^(@)C` to `67^(@)C` (Specific heat capacity of water is 1 cal `g^(-1@)C^(-1)`)

Explanation

Given, mass of water = 250 g
Change in temperature, `Deltat`=final temperature - initial temperature = `67^@C -27^@C = 40^@C`
Specific heat capacity of water = 1 cal `g^(-1)""^(@)C^(-1)`
Heat energy supplied, `Q=MsDeltat`
`Q=250 xx 1 xx 40` = 10,000 cal = 10 kcal