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.
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 -------------------
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.
Specific heat for aluminium = 0.214 Heat required = 38.2 x 0.214 x (275 - 102) = 1414.24 calories
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 ============
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
Approx 4974 Joules.
0.131 joules/gram'C x 1.3 grams x (46-25)'C = 3.5763 joules
21 Kg = 2100 grams to rise the temperature of this amount of water by 2 degrees Celsius you need 2*2100 = 4200 calories or 17572.8 Joules.
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 -------------------
q = [ 0.803 J/( g degrees Celsius)] ( 2 multiplied by '10 to the power of 6' grams) ( 19 degree Celsius) q = 3.1 multiplied by '10 to the power of 7' Joules
The unit for specific heat is Joules/g-Kelvin or it can be Joules/g-Celsius J= Joules g= Grams C= Celsius
q(joules) = mass * specific heat * change in temperature q = (500 grams H2O)(4.180 J/goC)(100o C - 20o C) = 1.7 X 105 joules ================add this much heat energy to the water
700
I assume you mean 30o Celsius. Use this formula.q(joules) = mass * specific heat * change in temperatureq = (15 grams water)(4.180 J/gC)(40o C - 30o C)= 627 joules==========( perhaps 630 joules to be in significant figures territory )
The change in temperature is 25 degrees Celsius, meaning it takes 22.48 joules per degree of change. The specific heat of iron is 0.449 J/g degree Celsius. This means that the mass of iron must be 50.07 grams
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 ===========================
It takes 4.186 Joules to heat one gram of water by 1-degree Celsius. 4.186 * 4000 = 16,744 Joules to heat 4 kilos of water by 1-degree. 16,744 * 70 = 1,172,080 Joules. The above assumes that one litre of water weighs exactly 1 Kilogram.