Question

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.

Answer 1

383.34 J/kg.K

Step-by-step 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