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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

User Botan
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1 Answer

3 votes

Answer:


\Delta T=194^oC

Step-by-step 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=(Q)/(mCp)=(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.

User Matt Victory
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