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An open container holds ice of mass 0.570 kg at a temperature of -17.2 ∘C . The mass of the container can be ignored. Heat is supplied to the container at the constant rate of 820 J/minute . The specific heat of ice to is 2100 J/kg⋅K and the heat of fusion for ice is 334×103J/kg.

Respuesta :

Answer:[tex]t=257.27\ min[/tex]

Explanation:

Given

mass of ice [tex]m=0.57\ kg[/tex]

Temperature [tex]T=-17.2^{\circ}C[/tex]

Heat supplied [tex]H=820\ J/min[/tex]

Heat of fusion of ice [tex]L=334\times 10^{3}\ J/kg[/tex]

assuming melting time to completely melt the ice is asked so

So heat supplied will melt the ice

[tex]H=mc\Delta T+mL[/tex]

[tex]820\times t=0.57\times 2100\times (17.2)+0.57\times 334\times 1000[/tex]

[tex]820t=210,968.4[/tex]

[tex]t=\dfrac{210968.4}{820}[/tex]

[tex]t=257.27\ min[/tex]

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