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, in terms of the heat, mass, heat capacity and change in temperature, we can analyze thermal changes as:

$Q=mCp\Delta T$

In such a way, we compute the required change in temperature as shown below:

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

Such change in temperature is positive indicating an increase in the temperature as the involved heat is positive, in means that heat was added to increase the temperature.

Best regards.