Question

A Sheet of metal weighing 750 g absorbs 15,650 J of energy when its temperature increases from 25°C to 250°C what is the specific heat of the metal

Answer 1

`C=0.152(J)/(g\°C)`

Step-by-step explanation:

Hello!

In this case, when doing calorimetry problems, we need to use the following equation:

`Q=mC(T_2-T_1)`

Whereas Q is the involved heat, m the mass, C the specific heat and T2 and T1 the final and initial temperatures respectively. Thus, since we need the specific heat, we proceed as follows:

`C=(Q)/(m(T_2-T_1))`

Thus, we plug in to obtain:

`C=(15,650J)/(750g(250\°C-25\°C)) \\\\C=0.152(J)/(g\°C)`

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