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 temperature of the room

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