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Determine the quantity of heat transferred by 20.0 grams of water where the temperature of the water changes from 65°C to 20°C, the temperat
Question
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Answers ( )
Answer:
3780J
Explanation:
Data obtained from the question include:
M (mass) = 20g
T1 (initial temperature) = 65°C
T2 (final temperature) = 20°C
ΔT (change in temperature) =
T1 — T2 (since we are cooling)
ΔT = 65 — 20 = 45°C
C (specific heat capacity) = 4.2J/g°C
Q (heat) =?
Q = MCΔT
Q = 20 x 4.2 x 45
Q = 3780J
Therefore, the heat transferred is 3780J