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