Question

Suppose that a 60.0g of water at 23.52°C was cooled by the removal of 813J of heat, calculate the change in temperature using the specific heat equation

Answer 1

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Answer 2

Answer:

Heat= ms(T2-T1)

where m= mass of substance, s= specific heat of substance, T2 and T1 are final and initial temperatures

813/4.2= 60*1*(T2-T1) ( checking units, we need to convert heat from joule to calorie)

Change in temperature is therefore 3.22 K.