Question

The amount of heat required in converting 1 gram ice at _10 degree into steam at 100degree celsius will be.

Answer 1

amount of heat required = 3.0334 kJ

mass of ice , m = 1g

temperature of ice , Ti = -10°C

heat required to increase the temperature of ice upto 0°C , h1 = ms∆T

= 1g × 0.5cal/g.°C × {0°C -(-10°C)}

= 1g × 0.5 × 10

= 5 cal

heat required to melt ice of mass 1g, h2 = ml_f

= 1g × 80 cal/g

= 80 cal

now increase the temperature of water to 100°C , h3 = ms'∆T'

= 1g × 1 cal/g.°C × 100°C

= 100 cal

heat required to convert water into steam, h4 = ml_v

= 1g × 540 cal/g

= 540 cal

so, total heat required = h1 + h2 + h3 + h4

= 5cal + 80cal + 100 cal + 540 cal

= 725 cal

we know, 1Cal = 4.184 Joule

so, heat = 725 × 4.184 Joule

= 3033.4 Joule

= 3.0334 kJ

also read similar questions: Heat required to convert one gram of ice at 0oc into steam at 100o c is

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Answer 2

$\huge\bold\purple{Answer:-}$

mass of ice , m = 1g

temperature of ice , Ti = -10°C

heat required to increase the temperature of ice upto 0°C , h1 = ms∆T

= 1g × 0.5cal/g.°C × {0°C -(-10°C)}

= 1g × 0.5 × 10

= 5 cal

heat required to melt ice of mass 1g, h2 = ml_f

= 1g × 80 cal/g

= 80 cal

now increase the temperature of water to 100°C , h3 = ms'∆T'

= 1g × 1 cal/g.°C × 100°C

= 100 cal

heat required to convert water into steam, h4 = ml_v

= 1g × 540 cal/g

= 540 cal

so, total heat required = h1 + h2 + h3 + h4

= 5cal + 80cal + 100 cal + 540 cal

= 725 cal

we know, 1Cal = 4.184 Joule

so, heat = 725 × 4.184 Joule

= 3033.4 Joule

= 3.0334 kJ