Question

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

Answer 1

Answer:

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

Answer 2

Answer:

The correct answer is 1.194 J/g.ºC

Explanation:

The heat released by the material is absorbed by the water. We put a minus sign (-) for a released heat and a plus sign (+) for an absorbed heat.

We know the mass of the material (mass mat= 25.0 g) and the mass of water (mass H20= 100.0 g) and the specific heat capacity of water is known (Shw=4.18 J/g.ºC), so we can equal the heat released by the material and the heat absorbed by water y calculate the specific heat capacity of the material (Shm) as follows:

heat released by material = heat absorbed by water

-(mass material x Shm x ΔT)= mass water x Shw x ΔT

-(25.0 g x Shm x (24ºC - 80ºC)= 100.0 g x 4.18 J/g.ºC  x (24ºC-20ºC)

25.0 g x Shm x (56ºC) =  100.0 g x 4.18 J/g.ºC x 4ºC

⇒Shm= (100.0 g x 4.18 J/g.ºC x 4ºC)/(25.0 g x 56ºC)

 Shm= 1.194 J/g.ºC