Question

What is the temperature change on a 75.0 gram sample of mercury if 480.0 cal of heat are added to it? The specific heat of mercury = 0.033 cal/g C

Answer 1

Answer:

$\Delta T=194^oC$

Explanation:

Hello,

In this case, the relationship between the mass, heat, temperature and heat capacity id given by:

$Q=mCp\Delta T$

In such a way, the temperature change results:

$\Delta T=\frac{Q}{mCp}=\frac{480.0cal}{75.0g*0.033cal/(g^oC)} \\\\\Delta T=194^oC$

Clearly it is a positive change which means the temperature increases as heat is added.

Regards.