Step-by-step explanation:
A) Heat absorbed by water (Joule)
We can find the heat absorbed by water using this equation:
Q = m * C * ΔT
Where Q is the heat, m is the mass of water, C is the specific heat of water and ΔT is the change in temperature.
We have 3 sets of values, we can get one heat for each trial and then average them.
Trial 1:
m = 9.007 g
C = 4.184 J/(g*C)
ΔT = Tfinal - Tinitial = 31.3 °C - 22.2 °C
ΔT = 9.1 °C
Q1 = m * C * ΔT
Q1 = 9.007 g * 4.184 J/(g*C) * 9.1 °C
Q1 = 342.94 J
Trial 2:
m = 8.811 g
C = 4.184 J/(g*C)
ΔT = Tfinal - Tinitial = 34.0 °C - 22.4 °C
ΔT = 11.6 °C
Q2 = m * C * ΔT
Q2 = 8.811 g * 4.184 J/(g*C) * 11.6 °C
Q2 = 427.64 J
Trial 3:
m = 8.826 g
C = 4.184 J/(g*C)
ΔT = Tfinal - Tinitial = 33.2 °C - 23.7 °C
ΔT = 9.5 °C
Q3 = m * C * ΔT
Q3 = 8.826 g * 4.184 J/(g*C) * 9.5 °C
Q3 = 350.82 J
Average:
Q = (Q1 + Q2 + Q3)/3
Q = (342.94 J + 427.64 J + 350.82 J)/3
Q = 373.8 J ---> Heat absorbed by water
B) Heat lost by metal (Joule):
We placed a hot piece of metal inside the water. The piece of metal got cooled and the water was heated by the metal.
If we neglect the heat that was lost to the surroundings, the amount of heat that the water absorbed is equal to the amount of heat lost by the metal (with different sign).
Qmetal = - Qwater
Qmetal = - 373.8 J
C) Specific heat of the metal (J/g °C)
We can also use the same equation that we used for the water.
Qmetal = m * Cmetal * ΔTmetal
We will have to solve the equation for Cmetal.
Cmetal = Q metal /(m * ΔTmetal)
The average mass of the piece of metal is:
m = 8.881 g
The initial temperature of the piece of metal is:
Tinitial = 100 °C
Trial 1:
Qmetal1 = - 342.94 J
ΔT = Tfinal - Tinitial = 31.3 °C - 100 °C
ΔT = - 68.7 °C
Cmetal1 = Qmetal1 /(m * ΔTmetal)
Cmetal1 = -342.94 J/(8.881 g * (-68.7 °C)
Cmetal1 = 0.562 J/(g°C)
Trial 2:
Qmetal2 = - 427.64 J
ΔT = Tfinal - Tinitial = 34.0 °C - 100 °C
ΔT = - 66.0 °C
Cmetal2 = Qmetal2 /(m * ΔTmetal)
Cmetal2 = -427.64 J/(8.881 g * (-66.0 °C)
Cmetal2 = 0.730 J/(g°C)
Trial 3:
Qmetal3 = - 427.64 J
ΔT = Tfinal - Tinitial = 33.2 °C - 100 °C
ΔT = - 66.8 °C
Cmetal3 = Qmetal3 /(m * ΔTmetal)
Cmetal3 = -350.82 J/(8.881 g * (-66.8 °C)
Cmetal3 = 0.591 J/(g°C)
Average:
Cmetal = (Cmetal1 + Cmetal2 + Cmetal3)/3
Cmetal = (0.562 J/(g°C) + 0.730 J/(g°C) + 0.591 J/(g°C))/3
Cmetal = 0.628 J/(g°C)
Answers:
a) The water absorbed 373.8 J
b) The heat lost by the metal is -373.8 J
c) The specific heat of the metal is 0.628 J/(g°C)