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The specific heat of ice is 0.492 cal/ (g × °c). how many calories of heat are required to raise 100.0 g of ice from -20.0°c to -0.5°c?

Respuesta :

carlosego

For this case we have that the number of calories is given by the following equation:

C = mCpΔt

Where,

m: mass

Cp: specific heat

Δt : temperature differential

Substituting values we have:

[tex] C = (100) * (0.492) * (- 0.5 - (- 20))

C = 959.4
[/tex]

Answer:

Around 959.4 calories of heat are required to raise 100.0 g of ice from -20.0 ° c to -0.5 ° c

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