A 38.4-g sample of ethylene glycol, a car radiator coolant, loses 852 J of heat. What was the initial temperature of the ethylene glycol if the final temperature is 32.5°C? (c of ethylene glycol = 2.42 J/g·K)

296.33K

Explanation:

Data;

Q = 852J

mass (m) = 38.4g

θ₂ = 32.5°C = 305.5K

θ₁ = ?

c = 2.42 J/g. K

From the equation of heat transfer;

Heat transfer (Q) = MC∇θ

Q = mc (θ₂ - θ₁)

852 = 38.4 * 2.42 * (305.5 - θ₁)

852 = 28389.504 - 92.928θ₁

Collect like-terms

92.928θ₁ = 27537.504

Divide both sides by 92.928 to make θ₁ the subject of formula.

θ₁ = 296.33K