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Heat

200g of a col...

Question

$200$ g of a cold ball at $20^{o}$ C is dropped in an equal amount of water at $80^{o}$ C. The resulting temperature is $60^{o}$ C. This means that specific heat of solid is.

One fourth of water

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One half of water

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Twice of water

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Four times of water

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Solution

The correct option is B One half of water
Heat gain = Heat lost
$200 \times S_{b a l l} \times (60 - 20) = 200 \times S_{w a t e r} \times (80 - 60) \Rightarrow S_{b a l l} = \frac{20}{40} S_{w a t e r} = \frac{1}{2} S_{w a t e r} \Rightarrow S_{b a l l} = \frac{1}{2} S_{w a t e r}$

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200g of a cold ball at 20oC is dropped in an equal amount of water at 80oC. The resulting temperature is 60oC. This means that specific heat of solid is.

The correct option is B One half of waterHeat gain = Heat lost200×Sball×(60−20)=200×Swater×(80−60)⇒Sball=2040Swater=12Swater⇒Sball=12Swater

$200$ g of a cold ball at $20^{o}$ C is dropped in an equal amount of water at $80^{o}$ C. The resulting temperature is $60^{o}$ C. This means that specific heat of solid is.

The correct option is B One half of water
Heat gain = Heat lost
$200 \times S_{b a l l} \times (60 - 20) = 200 \times S_{w a t e r} \times (80 - 60) \Rightarrow S_{b a l l} = \frac{20}{40} S_{w a t e r} = \frac{1}{2} S_{w a t e r} \Rightarrow S_{b a l l} = \frac{1}{2} S_{w a t e r}$