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# Heat required to melt 1 g of ice is 80 cal. A man melts 60 g of ice by chewing in 60 seconds. What is his power?A) 1.33 WB) 0.75 WC) 336 WD) 4800 W

Last updated date: 20th Jun 2024
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Hint:The heat required to melt one gram of solid to liquid is known as the latent heat of fusion. Total heat required is mass multiplied by latent heat of fusion.

Complete step by step solution:
We are given that 1 g of ice melts when it is provided with 80 cal of heat. This is the latent heat of fusion of ice to water. But calorie is not the SI unit of heat so we first need to convert cal to the SI unit of heat which is joule (J). We represent latent heat of fusion by the letter L.
L = 80 cal
To convert cal to joule, we need to multiply it by 4.2, because 1 cal = 4.2 J. On converting cal to joule, we obtain,
L = 80 cal = 80$\times$4.2 J = 336 J
This heat is the heat required to melt 1 g of ice. To find the total heat required to melt 60 g of ice, we need to multiply the latent heat of fusion with the mass of ice to be melted, which is equal to 60 g. We now find the total heat required, Q,
Q = mL = 60$\times$336 J = 20160 J
But we need to find the power used by the man in melting the ice in 60 seconds. For that, we find the heat given in one second, which can be obtained by dividing the total heat by the time taken by the ice to melt.
So, power used by the ice to melt, P = $\dfrac{Q}{t} = \dfrac{{20160}}{{60}}$ = 336 W
Hence, power delivered by man to melt the ice is 336 W.

Therefore, the correct answer is option C.

Note:We can also use the direct expression of total heat without performing the calculation to find it in the expression of power. On doing that, one step of calculation would be reduced.