The Starship Enterprise is caught in a time warp and Spock is forced to use the primitive techniques of the 20th century to determine the specific heat capacity of an unknown mineral. The 128-g sample was heated to 96.2°C and placed into a calorimeter containing 77.2 g of water at 20.0°C. The heat capacity of the calorimeter was 13.9 J/K. The final temperature in the calorimeter was 23.3°C. What is the specific heat capacity (in J/g°C) of the mineral? Enter to 4 decimal places.

The heat released by the unknown mineral ([tex]-Q_u[/tex], where we will use the minus sign to indicate release of heat) is the heat absorbed by the calorimeter with water [tex]Q_w[/tex], so we can write:

[tex]-Q_u=Q_w[/tex]

The formula that relates heat with the mass of a material, its specific heat capacity and the temperature difference it suffers because of that heat is:

[tex]Q=mc \Delta T=mc(T_f-T_i)[/tex]

So we substitute this into the previous equation to get:

[tex]-m_uc_u(T_{uf}-T_{ui})=m_wc_w(T_{wf}-T_{wi})[/tex]

Since at the end the final temperatures will be the same, namely

[tex]T_{uf}=T_{wf}=T_{f}[/tex]

We can write

[tex]-m_uc_u(T_f-T_{ui})=m_wc_w(T_f-T_{wi})[/tex]

Since what we want is the specific heat capacity of the unknown mineral [tex]c_u[/tex], we can write:

[tex]c_u=-\frac{m_wc_w(T_f-T_{wi})}{m_u(T_f-T_{ui})}[/tex]

In this particular equation it is not necessary to transform g to Kg (since the units of masses are cancelling out) or Celsius to Kelvin (since temperature is only appearing as a difference), but if one is not sure, the units should be transformed to S.I. since Joules is in S.I.

Finally then we have:

[tex]c_u=-\frac{(77.2g)(13.9J/K)(23.3C-20.0C)}{128g(23.3C-96.2C)}=0.37949717078J/K[/tex]

Since it asks for 4 decimal places, this must be written as:

[tex]c_u=0.3795J/K[/tex]