Calculate the quantity of heat needed to heat 8.0 grams of water from 42.0 OC to 45.0 OC.

Answer 1

Hi Angie

The answer to this question lies in the definition of specific heat. Specific heat of water is 1 Calorie/gm-^oC.

Energy necessary to raise the temperature of 1 gm of water by 1^oC. = 1 calorie (specific-heat definition)

Energy necessary to raise the temperature of 8 gm of water by 1^oC = 8 x 1 calorie

Energy necessary to raise the temperature of 8 gm of water by 3^oC = 3 x 8 x 1 calories

If you follow this logic, you can do similar problems without any difficulty.