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

The correct option is A 10 g Let the required mass of ice be m . m× 80+m× 1× (20 0)=100× 1× (30 20) ⇒ 80m+20m=1000 ⇒ m= 1000 100 =10g ...

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Byju's Answer

Standard XII

Chemistry

Heat Capacity

How much ice ...

Question

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

10 g

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Solution

The correct option is A $10 g$
Let the required mass of ice be $m$ .
$m \times 80 + m \times 1 \times (20 - 0) = 100 \times 1 \times (30 - 20) \Rightarrow 80 m + 20 m = 1000 \Rightarrow m = \frac{1000}{100} = 10 g$

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

The correct option is A 10gLet the required mass of ice be m.m×80+m×1×(20−0)=100×1×(30−20)⇒80m+20m=1000⇒m=1000100=10g

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

The correct option is A $10 g$
Let the required mass of ice be $m$ .
$m \times 80 + m \times 1 \times (20 - 0) = 100 \times 1 \times (30 - 20) \Rightarrow 80 m + 20 m = 1000 \Rightarrow m = \frac{1000}{100} = 10 g$