Calculate the heat energy required to raise the temperature of 2 kg of water from `10^@C` to `50^@C`. Specific heat capacity of water is `4200 JKg^(-1) K^(-1)`

Explanation

Given m=2kg , `DeltaT`=(50-10)=`40^@C`
In terms of Kelvin, `DeltaT`=(323.15 - 283.15)=40 K, `C=4200 J Kg^(-1) K^(-1)`
`therefore` Heat energy required , `Q=mxxCxxDeltaT=2xx4200xx40`=3,36,000 J