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We guarantee that you'll never find yourself in a situation where you're expected

to do that job. But if you are, you're sure to pull it off in a flash, because it requires

no energy.

You'll never find any ice at 40° C ... at least not in a place where you can last

long enough to look at it. Ice that was brought to a place at 40° C has melted,

a long time ago.

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Q: How many joules of heat are required to melt 40 grams of ice at 40 degrees Celsius?
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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 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 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 many joules are absorbed by the water with a mass of 8 kg in order to raise the temperature from 20 degrees to 70 degrees?

q(joules) = mass * specific heat * change in temperature ( 8 kg = 8000 grams ) q = (8000 grams H2O)(4.180 J/gC)(70o C - 20o C) = 1.7 X 106 joules ============


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

Related questions

The specific heat of gold is 0.131 Joules per gram. Celsius How much energy is required to heat 1.3 grams of gold from 25 Celsius to 46 Celsius?

The change in temperature is 21 degrees Celsius. To calculate the energy required, we use the formula: Energy = mass * specific heat * change in temperature. Plugging in the values, Energy = 1.3g * 0.131 J/g°C * 21°C = 35.247 Joules. Therefore, 35.247 Joules of energy is required to heat 1.3 grams of gold from 25°C to 46°C.


What is the total heat loss if you have 18 grams of aluminum and the temperature changes from 64 degrees Celsius to 48 degrees Celsius?

Approx 4974 Joules.


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 much energy is required to raise 21kg of water by 2 degrees Celsius?

The specific heat capacity of water is 4.18 J/g°C. To calculate the energy required to raise 21 kg of water by 2 degrees Celsius, use the formula: Energy = mass x specific heat capacity x temperature change. Plugging in the values, the energy required is 21,084 Joules.


What is the measurement for specific heat?

The unit for specific heat is Joules/g-Kelvin or it can be Joules/g-Celsius J= Joules g= Grams C= Celsius


How many kilojoules of energy are necessary to raise the temperature of 3 kilograms of cast iron from 30 degrees celsius to 120 degrees celsius?

I will use this formula. Some conversion will be required. ( I only know specific heat iron in J/gC ) q(Joules) = mass * specific heat * change in temperature Celsius 3 kilograms cast iron = 3000 grams q = (3000 g)(0.46 J/gC)(120 C - 30 C) = 124200 Joules (1 kilojoule/1000 joules) = 124.2 kilojoules of energy needed ===========================


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 many joules of heat energy are required to raise the temperature of 6 grams of copper of 100 degrees Celsius to 150 degrees Celsius?

The specific heat capacity of copper is 0.385 J/g°C. To calculate the heat energy required, you use the formula: Q = mcΔT, where Q is the heat energy, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. Plugging the values in, you get Q = 6g * 0.385 J/g°C * (150°C - 100°C) = 92.4 Joules.


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 must be lost by 2 grams of water at 0 degrees Celsius to change it to ice?

The heat lost by water at 0 degrees Celsius to change to ice is equal to the heat of fusion of water, which is about 334 joules per gram. So, for 2 grams of water, the heat loss would be 2 * 334 = 668 joules.


How many kilo joules are need to raise the temperature 17 Celsius to 34 Celsius?

To calculate the energy required, you can use the formula: Q = mcΔT Where Q is the energy in joules, m is the mass in grams, c is the specific heat capacity, and ΔT is the change in temperature in Celsius. Given the temperature change is from 17°C to 34°C and you have the mass and specific heat capacity, you can calculate the energy required in kilojoules.


How many joules of heat required to raise the temperature of 25 grams of h2o from 21 degrees Celsius to 35.5 degrees?

The specific heat capacity of water is 4.184 J/g°C. The change in temperature is 35.5°C - 21°C = 14.5°C. Using the formula Q = mcΔT, where Q is the heat energy, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature, you can calculate that the heat required is approximately 1554.7 joules.