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What is the heat in joules required to raise the temperature of 25 grams of 0^@C to 100^@C? What is the heat in Calories? (Specific heat of water =4.18 J//g^@C )

Answer»

SOLUTION :Given m=25 G , `DeltaT`=(100-0)=`100^@C`
or in terms of KELVIN (373.15-273.15)=100K, `C=4.18J//g^@C`
Heat energy required ,Q=m x C x `DeltaT` =25 x 4.18 x 100 = 10450 J


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