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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)

Answer»

Solution :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


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