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The temperature of a sample of water increases from 20 degrees Celsius to 46.6 degrees Celsius as it absorbs 5650 calories of heat. What is the mass of the sample

Respuesta :

Willchemistry
ΔT = 46.6 - 20

ΔT = 26.6ºC

Specific heat of water is 1.0 cal/g °C

Q = m * C * ΔT 

5650 = m * 1.0 * 26.6

5650 = m * 26.6

m = 5650 / 26.6

m = 212.4 g

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