How many calories must be absorbed by 20.0 g of water to increase its temperature from 383.0 C to 303.0 C

1600 calories

Explanation:

Data obtained from the question include:

M (mass of water) = 20g

T1 (initial temperature) = 383°C

T2 (final temperature) = 303°C

ΔT (change in temperature) = T1 - T2 = 383 - 303 = 80°C

C (specific heat capacity of water) = 1 calorie/g°C

Q (heat) =?

Using the the equation Q = MCΔT, the heat in calories absorbed by the water can be obtained as follow:

Q = MCΔT

Q = 20 x 1 x 80

Q = 1600 calories

Therefore, the heat absorbed by the water is 1600 calories

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