Answer:
T₂ ≈ 107.85∘C
Step-by-step explanation:
The question didn't state if the volume is constant or not as such, we can apply the first law of thermodynamic
From the first law of thermodynamic,
ΔU = Q - W
where ΔU = Internal Energy, Q = Quantity of heat absorbed, W = Amount of work done.
Q = 1200 J and W = 2020 J
∴ ΔU = 1200 -2020 = -820 J.
Using the ideal gas equation,
ΔU = 3/2nRΔT...................................equation 1
where n = number of moles, R = Molar gas constant, ΔT = Change in temperature = (T₂ - T₁).
Modifying equation 1,
ΔU = 3/2nR(T₂ -T₁)...............................equation 2.
making T₂ the subject of the relation in equation 2,
T₂ = {2/3(ΔU)/nR}+T₁........................ equation 3
where T₁=121∘C, R= 8.314 J / mol, n=5 moles, ΔU=-820 J
Substituting these values into equation 3,
∴ T₂ ={ 2/3(-820)/(5×8.314)}+121
T₂ = {2×(-820)/ (3×5×8.314)}+121
T₂={-1640/124.71}+ 121
T₂ = {-13.151} + 121
∴T₂ = 121 - 13.151 = 107. 849∘C
T₂ ≈ 107.85∘C