An 11.98-gram sample of zinc metal is placed in a hot water bath and warmed to 78.4°C. It is then removed and placed into a Styrofoam cup containing 50.0 mL of room temperature water (T=27.0°C; density = 1.00 g/mL). The water warms to a temperature of 28.1°C. Determine the specific heat capacity of the zinc.

383.34 J/kg. K

Explanation:

From the question,

Heat lost by zinc = heat gained by water in the Styrofoam cup.

c'm' (t₁-t₃) = cm (t₃-t₂) ... Equation 1

Where c' = specific heat capacity of zinc, m' = mass of zinc, t₁ = initial temperature of zinc, t₂ = initial temperature of water in the Styrofoam cup, t₃ = temperature of the mixture

Make c' the subject of the equation

c' = cm (t₃-t₂) / m' (t₁-t₃) ... Equation 2

given: m' = 11.98 g = 0.01198 kg, m = density of water*volume of water = (1*50) = 50 g = 0.05 kg, t₁ = 78.4°C, t₂ = 27°C, t₃ = 28.1°C

Constant: c = 4200 J/kg. K

Substitute these values into equation 2

c' = 0.05*4200 (28.1-27) / [0.01198 * (78.4-28.1) ]

c' = 231/0.602594

c' = 383.34 J/kg. K