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Dahasolnce [82]
3 years ago
14

How much heat must be removed from 15.5 g of water at 90.0°C to cool it to 43.2°C?

Chemistry
1 answer:
Katyanochek1 [597]3 years ago
3 0

Answer:

Heat, Q = 3035.073 J

Explanation:

It is given that,

Mass of water, m = 15.5 g

Initial temperature, T_i=90^{\circ} C

Final temperature, T_f=43.2^{\circ}

The specific heat of water is, c = 4.184 J/ g°C

The heat removed or absorbed by water is given by formula as :

Q=mc\Delta T\\\\Q=mc(T_f-T_i)\\\\Q=15.5\times 4.184\times (43.2-90)\\\\Q=-3035.073\ J

So, the heat of 3035.073 J is removed from 15.5 g of water.

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