The final temperature when 150 mL of water at 90.0°C is added to 100.0 mL of water at 30.0°C is around ________.

Question

Quizplus · Accepted Answer

To solve this problem, we can use the formula Q = mcΔT, where Q is the heat transferred, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. Since we are assuming no heat is lost to the surroundings and the two substances are in thermal equilibrium (i.e. the final temperature is the same for both substances), we have:
Q₁ = Q₂
m₁c₁ΔT₁ = m₂c₂ΔT₂
where the subscripts 1 and 2 refer to the hot and cold water, respectively. Solving for ΔT₂:
ΔT₂ = (m₁c₁ΔT₁) / (m₂c₂)
Since we are only interested in the final temperature, we don't need to calculate the heat transferred (Q) explicitly. Instead, we can use the fact that the sum of the masses is conserved:
m₁ + m₂ = 150 mL + 100 mL = 250 mL
Converting to grams:
m₁ + m₂ = 250 g
Using the specific heat capacity of water (c = 4.184 J/g°C) and substituting the values we know:
ΔT₂ = (150 g)(4.184 J/g°C)(90.0°C - T₂) / (100 g)(4.184 J/g°C)
Simplifying:
3.522(90.0 - T₂) = 3.962T₂
316.98 - 3.522T₂ = 3.962T₂
316.98 = 7.484T₂
T₂ ≈ 42.4°C
Therefore, the answer is D) 60 °C: the final temperature is somewhere between the initial temperatures of the two substances, and the only answer choice within that range is 60 °C.

The Final Temperature When 150 ML of Water at 90