Answered

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How much heat (in kJ) would need to be removed to cool 150.3 g of water from 25.60°C to -10.70°C?

Sagot :

Answer:

Q = -22.9 kJ

Explanation:

Given that,

Mass of water, m = 150.3 g

Water gets cool from 25.60°C to -10.70°C.

The specific heat of water, c = 4.2 J/g°C

The formula for heat needed is given by :

[tex]Q=mc\Delta T\\\\Q=150.3\times 4.2 \times (-10.7-25.6)\\\\Q=-22914.738\\\\or\\\\Q=22.9\ kJ[/tex]

So, 22.9 kJ of heat is needed to be removed to cool.

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