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If the quantity of heat lost or gained is (ΔH) = mass × change in T × specific heat, what is the amount of heat required to raise 200.0 g of water from 22.00°C to 100.0°C? The specific heat of water is 4.180 Joules per gram-°C.

Respuesta :

Edufirst
1) formula:


ΔH = m * Cs * ΔT


2) data


m = 200.0 g

ΔT = 100.0 °C - 22.00°C

Cs = 4.180 J / (g * °C)



3) Calculations:


ΔH = 200 g * (100.0°C - 22.00°C) * 4.180 J / (g * °C) = 65,208 J



Answer: 65,208 J

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