If 585.24 Joules of heat are added to 53.2 grams of water at 24.15oC, what will the new temp. be?

1 answer:

3 0

E = mct
Energy = (mass) x (specific heat capacity of water) x (change in temp)

585.24 = 53.2 x 4.2 x (X-24.15)

585.24 divided by 53.2 divided by 4.2 = X - 24.15

2.62 = X - 24.15

X= 26.77degrees C

(Specific heat capacity for water is 4.2 but is different for other liquids)

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