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28-12-2021
Physics

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what is q if 28.6 g of water is heated from 22.0°c to 78.3°c? the specific heat of water is 4.184 j/g·°c.

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28-12-2021

Answer:

6736 J

Explanation:

Knowing that Q = cmΔT:

Q = 4.184 J/(g·°C) · 28.6g · (78.3 °C - 22.0 °C) = 6736 J.

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