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All calculations below are assumed done with all materials (ice, water, and steam) being at normal atmospheric pressure.

3.0kg is 3000 gm. Heat required to raise temperature of ice from -10 degree celcius to 0 degrees celcius is 3000 x 10 cals. (A)

The latent heat of fusion of ice is 80 cals per gm. So heat needed to melt 3000 gm of ice at 0 degrees celcius is 3000 x 80 cal. (B)

The heat required to convert 3000 gms of water from 0 degrees celcius to 100 degrees celcius is 3000 x 100 cals. (C)

The latent heat of vaporization of steam is approximately 540 cals per gm. Hence heat required to convert 3000 gms of water to steam at 100 degrees celcius is 3000 x 540 cals. (D)

Now add (A) + (B) + (C) + (D) = 30000 + 240000 + 300000 + 1620000 cals.

And that is = 2190000 cals = 2.19 X 10 to power 6 cals.

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Q: How much heat is needed to change 3.0kg of ice at -10 degree into steam at 300 degree Celsius?
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How many calories are needed to change 1 gram of 100 degrees Celsius boiling water to 100 degrees Celsius steam?

This is the latent heat of vaporisation of water, which at standard pressure, is 539 calories (per gram).


How much heat energy is required to change a block of ice with a mass of 55.6 grams at -11.9 degrees Celsius to steam at 123 degree Celsius?

For one gram of ice, it takes 11.9 calories to change the temperature to 0°, 80 calories to melt the ice, 100 calories to raise the water temperature to 100°, 540 calories to change the water to steam, and 23 calories to raise the steam temperature to 123°. That's a total of (11.9 + 80 + 100 + 540 + 23) calories or 754.9 calories. So to do the same to 55.6 grams of ice requires 55.6 times as much heat. 754.9 calories times 55.6 equals approximately 41972 calories (about 42 kilocalories).


How much heat energy will be required to lower the temperature of 2.67kg of steam from 282 degrees Celsius to 105 degrees Celsius?

You mean how much heat energy will be lost/transferred as you are losing Joules here. All in steam, so a simple q problem and no change of state. 2.67 kg = 2670 grams q = (2670 grams steam)(2.0 J/gC)(105 C - 282 C) = - 9.45 X 105 Joules ----------------------------------- This much heat energy must be lost to lower the temperature of the steam.


Is the temp of steam at 100 degrees celsius same as 100 degrees celsius water?

Leaving aside the effects of pressure, yes, the temperatures are the same. But the amount of heat (thermal energy) per gram, is much greater for the steam.


How much heat is necessary to change a 52.0 gram sample of water at 33.0 degrees Celsius into steam at 110 degrees celsius?

Raise the temp of 52 grams of water from 33.0 C to 100 C = 52*67*4.184 = 14.577 kJConvert evaporate 52 g of water to steam without change of temp = 52*2259 = 117.468 kJRaise the temp of 52 grams of steam from 100 C to 110 C = 52*10*2.02 = 1.051 kJTotal energy required = 133.095 kJ = 31,811 calories or 31.811 kCal.

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