Find the result of mixing 10gm of ice at -8°C with 10gm of water 12°C? (Sp. heat of ice=0.5cal/gm°c, latent heat of fusion of ice=80cal/gm)
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Since ice melted at 0 Celesius then the heat gained by ice (latent heat) will melt it so you should substitute in that law
Q=mlfQ=mlf ..where Q is the heat required to convert ice to water , m is the mass of ice and lf is the latent heat of fusion
Q=5∗80=400
calQ=5∗80=400cal
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