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Specific Heat Capacity

The amount of...

Question

The amount of heat required to convert 5 gm of ice at $- 20^{0} C$ to water at $100^{0} C$ is

450

cal

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950

cal

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100

cal

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none of these

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Solution

The correct option is B $950$ cal
Heat supply $= Q_{i c e_{- 20 \to 0}} + Q_{i c e \to w a t e r} + Q_{w a t e r_{0 \to 100}}$
Heat $= m s_{i c e} (0 - (- 20)) + m l_{f} + m s_{w a t e r} (100 - 0)$
$= 5 \times \frac{1}{2} \times 20 + 5 \times 80 + 5 \times 1 \times 100$
$= 50 + 400 + 500$
$= 950$ cal.

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Similar questions

0^{\circ}

20^{\circ}

g^{- 1 o} C^{- 1}

- 10^{0} C

100

^{\circ} C

80 C a l / g m,

540 C a l / g m

1 C a l / g m / C

5 g

0^{o} C

40^{o} C

= 80 c a l / g

= 1 c a l / g

0^{_{-}^{0}} C

100^{_{-}^{0}} C

L_{s t e a m}

Heat required to convert one gram of ice at $0^{\circ} C$ into steam at $100^{\circ} C$ is (given $L_{s t r e a m}$ = 536 cal/gm)

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