Question

A sample of iron, which has a specific heat capacity of 0.449 Jg^-1℃^-1, is put into a calorimeter (see sketch at right) that contains 100.0 g of water. The iron sample starts off at container 93.3 °C and the temperature of the water starts off at 22.0 °C. When the temperature of the water stops changing it's 25.7 °C. The pressure remains constant at 1 atm.

Required:
Calculate the mass of the iron sample. Be sure your answer is rounded to 2 significant digits.

Answer 1

`m_(iron)=32.1g`

Step-by-step explanation:

Hello!

In this case, since the interaction between hot iron and cold water allows the heat transfer from iron to water and therefore we can write up the following energetic equation:

`Q_(iron)+Q_(water)=0`

Whereas the heat terms can be written in terms of mass, specific heat and temperature change:

`m_(iron)C_(iron)(T_f-T_(iron)) + m_(water)C_(water)(T_f-T_(water)) = 0`

So we solve for the mass of iron as follows:

`m_(iron) = (m_(water)C_(water)(T_f-T_(water)))/(C_(iron)(T_f-T_(iron)) )`

Now, we plug in the given data to obtain:

`m_(iron) = (-100g*0.449(J)/(g\°C) (25.7\°C-22.0\°C))/(0.449(J)/(g\°C) (25.7\°C-93.3\°C) )`

`m_(iron)=32.1g`

Best regards!