A48 kg sample of water absorbs 366 kj of heat. the water was initially at 20.3 ∘c, what is its final temperature?

The final temperature of water = 22.1224 °C

Explanation:

Step 1 : defien the formula for heat transfer

q = m*c*ΔT

with q = heat transfer

with m = mass of the substance

with c = specific heat ( of water in this case)

with ΔT = difference between final temperature and initial temperature

In this case :

q=366kJ = 366 *10^3 J

mass of water = 48kg = 48000g

c= specific heat of wwater = 4.184 J / g °C

initial temperature = 20.3 °C

final temperature =  TO BE DETERMINED

Step 2: Calculate temperature

366000 J = 48000g * 4.184J/g °C * (T2 - 20.3)

366000 / (480000 * 4.184) = 1.8224 = (T2 - 20.3)

T2 = 20.3 + 1.82 =22.1224 ° C

The final temperature of water = 22.1224 °C

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