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The idea here is to:

* Look up the specific heat of water.

* Multiply the mass, times the temperature difference, times the specific heat of water.

You may need to do some unit conversions first; specifically, if the specific heat is given per kilogram, you can convert the grams to kilograms.

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Q: How much heat is necessary to change 30 grams of water at 40 degrees celsius into water at 60 degrees celsius?
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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.


How much heat is required to raise the temperature to 5.0 mL of water from 2.50 degrees Celsius to 75 degrees Celsius?

Density = grams/ml 1.00 g/ml = X g/5.0 ml = 5.0 grams water ============== q(joules) = mass * specific heat * change in temp. q = (5.0 grams)(4.180 J/gC)(75 C - 2.50 C) = 1515.25 Joules ---------------------------------/4.184 = 362 calories -------------------


How many calories are required to heat 38.2 grams of aluminum from 102 degrees celsius to 275 degrees celsius?

Specific heat for aluminium = 0.214 Heat required = 38.2 x 0.214 x (275 - 102) = 1414.24 calories


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.


How much heat is required to vaporize 10 pound ice?

If the ice starts at 0 degrees Celsius and vaporization takes place at 100 degrees Celsius, then ... 10 lbs = 4540 grams Q = 80(4540) + 1(100 - 0)(4540) + 540(4540) Q = 720(4540) = 3,268,800 calories

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