A piece of newly synthesized material of mass 25.0 g at 80.0C is placed in a calorimeter containing 100.0 g of water at 20.0C. If the final temperature of the system is 24.0C, what is the specific heat capacity of the material

1.2 J/g.°C or 1200 J/kg.°C

Explanation:

Heat lost by the material = heat gained by water.

CM(t₁-t₃) = cm(t₃-t₂) Equation 1

Where C = specific heat capacity of the material, M = mass of the material, c = specific heat capacity of water, m = mass of water, t₁ = initial temperature of the material, t₂ = initial temperature of water, t₃ = final temperature of the system.

make C the subject of the equation

C = cm(t₃-t₂)/M(t₁-t₃) Equation 2

Given: M = 25 g, m = 100 g, t₁ = 80 °C, t₂ = 20 °C, t₃ = 24 °C

Constant: c = 4.2 J/g.°C

Substitute into equation 2

C = 100(4.2)(24-20)/25(80-24)

C = 100(4.2)(4)/25(56)

C = 1680/1400

C =  1.2 J/g.°C or 1200 J/kg.°C

b. accuracy of the data matters alot more than its number of digits, organization or amount of detail. quantitative observation is more accepted than the qualitative observation.