Question

A 134.0 g sample of an unknown metal is heated to 91.0⁰C and then placed in 125 g of water at 25.0⁰C. The final temperature of the water is measured at 31.0⁰C. Calculate the specific heat capacity of the unknown metal. Specific Heat of water is 4.18 J/g*C pls answer as quickly as possible

Answer 1

`Cp_(metal)=39.0(J)/(g\°C)`

Step-by-step explanation:

Hello,

In this case, since the following equation relates heat, mass, specific heat and temperature:

`Q=mCp(T_2-T_1)`

For the two substances, we say that the heat lost by the metal equals the heat gained by the water, thus we have:

`m_(water)Cp_(water)(T_2-T_(water))=-m_(metal)Cp_(metal)(T_2-T_(metal))`

Thus, the specific heat of the metal turns out:

`Cp_(metal)=(m_(water)Cp_(water)(T_2-T_(water)))/(m_(metal)(T_2-T_(metal)))`

Therefore, we obtain:

`Cp_(metal)=(125g*4.18(J)/(g\°C) *(31.0-91.0)\°C)/(-134.0g*(31.0-25.0)\°C)\\\\Cp_(metal)=39.0(J)/(g\°C)`

Best regards.