In an experiment on the specific heat of a metal, a 0.30 kg block of the metal at 120 °C is dropped in a copper calorimeter (of water equivalent 0.025 kg) containing 100 cm³ of water at 27 °C. The final temperature is 35 °C. Calculate the specific heat of the metal?

Concept:

• In this question heat lost by the metal will be responsible for the rise in temperature of the calorimeter water.
• Thermal energy stored in a body can be calculated as:

Q = mc_pdT

Where m is the mass of the body, c_p is the heat capacity of the body at a constant temperature, dT is the temperature difference between the environment and the temperature of the body.

• But remember this heat totally cannot be extracted for work only the useful work obtained by Gibbs free energy.

Calculation:

Given: Mass of water m_w = 0.025 kg = 25 g ,

Mass of metal  M = 0.3 kg = 300g , T_i = 120^oC,

Final temperature, T_f  = 35^oC, 

Density of the water = 1 g/cm³

So, the mass of the water of 100 cm³ volume = 100 g

Specific heat of water, C_w = 4.186 Jg-1 K

⇒ Mc_pT_w = (M + m) × C_w× ( T_wi - T_f)

Heat lost by metal = Heat gained by the water and calorimeter system.

⇒ mCΔT_m = (M + m) × C_w × T_w

⇒ 300 × C × (120 - 35) = (100 + 25)× 4.186 × (35 -27)

⇒ C × 25500 = 10883.6

⇒ C = 0.164 J g^-1K^-1

Note: Here, the answer was not matching with the official options so we modified the option which was close to the correct answer.