Answer:
T_finalmix = 45 [°C].
Explanation:
In order to solve this problem, a thermal balance must be performed, where the heat is transferred from water with the highest temperature, to the water with the lowest temperature, so the final temperature of the mix must be equal once the thermal balance is achieved.
[tex]Q_{water}=Q_{methanol}[/tex]
where:
[tex]Q_{water}=m_{water}*Cp_{water}*(T_{waterinitial}-T_{final})[/tex]
mwater = mass of the water = 300 [g] = 0.3 [kg]
Cp_water = specific heat of the water = 4180 [J/kg*°C]
T_waterinitial = initial temperature of the water = 80 [°C]
T_finalmix = final temperature of the mix [°C]
[tex]Q_{water}=m_{water}*Cp_{water}*(T_{final}-T_{water})[/tex]
Now replacing:
[tex]0.3*4180*(80-T_{final})=0.3*4180*(T_{final}-10)\\100320-1254*T_{final}=1254*T_{final}-12540\\112860=2508*T_{final}\\T_{final}=45[C][/tex]
The final temperature of the mixture is 45 °C
Heat lost by warm water = Heat gained by cold water
Qᵥᵥ = Q꜀
MᵥᵥC(Tᵥᵥ – Tₑ) = M꜀C(Tₑ – T꜀)
0.3 × 4180 × (80 – Tₑ) = 0.3 × 4180 × (Tₑ – 10)
1254 (80 – Tₑ) = 1254(Tₑ – 10)
Clear bracket
100320 – 1254Tₑ = 1254Tₑ – 12540
Collect like terms
100320 + 12540 = 1254Tₑ + 1254Tₑ
112860 = 2508Tₑ
Divide both side by 2508
Tₑ = 112860 / 2508
Tₑ = 45 °C
Thus, the final temperature of the mixture is 45 °C
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