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2. 290 grams of water is to be heated from 24.0% to 100.0°C to make a cup of
tem, how much heat must be added? The specific heat of water is 4.18 J/g•C

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Answer:

92127.2 j

Explanation:

Given that :

Mass of water = 290 g

Specific heat of water (C) = 4.18 j/g.C

Initial Temperature (t1) = 24°C

Final temperature (t2) = 100°C

Using the relation :

Q = mCdt

dt = change in temperature =( 100 - 24)° = 76°C

Q = quantity of heat ; C = specific heat capacity;

m = mass of substance

Q = 290 g * 4.18 j/g.C * 76°C

Q = 92127.2 j

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