How much ice must be added to 100 g water at 30- C in order to reduce its temperature to 20- C

In a laboratory experiment for finding specific latent heat of ice, 100 g of water at 30^oC was taken in a calorimeter made of copper and of mass 10 g. When 10 g of ice at 0^oC was added to the mixture and kept within the liquid till the ice melted completely, the final temperature of the mixture was found to be 20^oC.

Calculate the value of the latent heat of fusion of ice from the data discussed above.

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How much ice must be added to 100 g water at 30∘C in order to reduce its temperature to 20∘C

The correct option is A 10gLet required amount of ice be mHeat gain = Heat lossm×80+m×1×20=100×1×(30−20)⇒m(80+20)=100×10⇒m=100×10100=10g

How much ice must be added to 100 g water at $30^{\circ} C$ in order to reduce its temperature to $20^{\circ} C$

The correct option is A $10 g$
Let required amount of ice be $m$
Heat gain = Heat loss
$m \times 80 + m \times 1 \times 20 = 100 \times 1 \times (30 - 20) \Rightarrow m (80 + 20) = 100 \times 10 \Rightarrow m = \frac{100 \times 10}{100} = 10 g$