Heat required to convert one gram of ice at 0- C into steam at 100- C is given Lstream - 536 cal-gm

Conversion of ice ( 0^∘ C ) into steam ( 100^∘ C ) is as follows Heat required in the given process = Q_1 + Q_2 + Q_3 = 1 × 80 + 1 × 1 × ( 100 0 ) + 1 × ...

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Standard XIII

Physics

Calorimetry

Heat required...

Question

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)

100 calorie

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0.01 kilo calorie

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716 calorie

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1 kilo calorie

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Solution

The correct option is C
716 calorie

Conversion of ice ( $0^{\circ} C$ ) into steam ( $100^{\circ} C$ ) is as follows

Heat required in the given process = $Q_{1} + Q_{2} + Q_{3}$

= $1 \times 80 + 1 \times 1 \times (100 - 0) + 1 \times 536 = 716 c a l$

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Similar questions

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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is (given

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

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)

Q. The amount of heat (in calories) required to convert 10g of ice at -

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is nearly:

(Specific heat of steam is

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)

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Heat required to convert one gram of ice at 0∘C into steam at 100∘C is (given Lstream = 536 cal/gm)

The correct option is C 716 calorie Conversion of ice ( 0∘C ) into steam ( 100∘C ) is as follows Heat required in the given process = Q1+Q2+Q3 = 1×80+1×1×(100−0)+1×536=716cal

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)

The correct option is C
716 calorie

Conversion of ice ( $0^{\circ} C$ ) into steam ( $100^{\circ} C$ ) is as follows

Heat required in the given process = $Q_{1} + Q_{2} + Q_{3}$

= $1 \times 80 + 1 \times 1 \times (100 - 0) + 1 \times 536 = 716 c a l$