howdy!
lets find the answer using the principle of conservation of energy. the equation for this is:
Q1 = Q2
where:
Q1 is the heat gained by the cooler substance (mass m1, specific heat C1, temperature change ΔT1).
Q2 is the heat lost by the warmer substance (mass m2, specific heat C2, temperature change ΔT2).
in this case, we have two portions of turpentine with different temperatures:
m1 = 94.8g, C1 = 1700 J/g°C, T1 = 15.1°C
m2 = 152g, C2 = 1700 J/g°C, T2 = 27.2°C
we need to find the final temperature T.f of the mixture.
first, calculate the heat gained and lost:
Q1 = m1 * C1 * ΔT1
Q2 = m2 * C2 * ΔT2
now, set Q1 = Q2 and solve for ΔT2 (the temperature change of the warmer substance):
ΔT2 = (m1 * C1 * ΔT1) / (m2 * C2)
plug in the values:
ΔT2 = (94.8g * 1700 J/g°C * (T.f - 15.1°C)) / (152g * 1700 J/g°C)
now, solve for T.f:
T.f - 15.1°C = (94.8g * 1700 J/g°C * (T.f - 15.1°C)) / (152g * 1700 J/g°C)
now, isolate T.f:
T.f - 15.1°C = (94.8 / 152) * (T.f - 15.1°C)
now, you can solve for T.f:
(1 / 1) = (94.8 / 152)
T.f - 15.1°C = 0.62368421
T.f = 15.1°C + 0.62368421
T.f ≈ 15.72°C
the final temperature of the mixture is 15.72°C.