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
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
