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If 4520 kJ of heat is needed to boil a sample of water, what is the mass of the sample of water (ΔHvaporiztion = 2260kJ/kg)?

If 4520 kJ of heat is needed to boil a sample of water, what is the mass of the sample of water (ΔHvaporiztion = 2260kJ/kg)?
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If 4520 kJ of heat is needed to boil a sample of water, what is the mass of the sample of water (ΔHvaporiztion = 2260kJ/kg)?

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User Sigute
by
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1 Answer

3 votes

Answer:2kg

Explanation:heat gained=heat lost

mass x 2260=4520

4530÷2260

And=2kg

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User Jonny Brooks
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