Question

What is the specific heat of tin if a mixture of 100.0 g of Cu at 10.0°C and 200.0 g of Sn at 120.0°C reach thermal equilibrium at a temperature of 69.0°C?

The specific heat of copper is 0.38 J/(g·°C).

Please see https://youtu.be/OoxQkc20szs for assistance.   J/(g ·°C)

Answer 1

The basic Calorimetry equation is;

                                                                        q = mCΔT

                                                          where     q = Heat energy required (in Joules)

                                                                        m = Mass (in Grams)

                                                                        C = Specific heat capacity (in J/ (g ·°C))

                                                                      ΔT = T_f - T_i (in°C)

                                                                            = Final temperature – Initial temperature (in°C)

Given that;

Mass of Cu                                    = 100 gm

Mass of Sn                                    = 200 gm

Initial temperature of Cu                = 10 °C

Final temperature of Cu                = 69 °C (Copper gains heat)

Initial temperature of Sn                = 120 °C

Final temperature of Sn                 = 69 °C (Tin loses heat)

             

At thermal equilibrium;

                                          q_Cu = -q_Sn

Therefore, (mCΔT) of Cu          = -(mCΔT) of Sn

100 x 0.38 x (69 – 10) = - (200 x C x (69 – 120))

2242                               = - (-10200 x C)

2242                               = 10200 x C

Therefore, C of Sn                    = 2242 ∕ 10200

                                          = 0.2198 J/ (g ·°C)