An unknown material, m1 = 0.49 kg, at a temperature of T1 = 92 degrees C is added to a Dewer (an insulated container) which contains m2 = 1.1 kg of water at T2 = 21 degrees C. Water has a specific heat of cw = 4186 J/(kg⋅K). After the system comes to equilibrium the final temperature is T = 31 degrees C. Detemine the specific heat of unknown material.

We know that heat relates to mass, specific heat and variation of temperature experimented because of this heat through the equation [tex]Q=mc\Delta T=mc(T_f-T_i)[/tex]. The heat released by the unknown material is absorbed by water, so we have [tex]Q_u=-Q_w[/tex], and we can write:

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

Since thermal equilibrium is reached we know that [tex]T_{cf}=T_{wf}=T_f=31^{\circ}C=304^{\circ}K[/tex], where we have added [tex]273^{\circ}[/tex] to convert the temperature from Celsius to Kelvin, as we must do. Since we want the specific heat of the unknown material, we do:

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

Which for our values is:

[tex]c_u=-\frac{(1.1kg)(4186J/kg^{\circ}K)((304^{\circ}K)-(294^{\circ}K))}{(0.49kg)((304^{\circ}K)-(365^{\circ}K))}=1540.5J/kg^{\circ}K[/tex]