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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
200 grams
700
69
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.
4.18J/g degrees C * 7.40 g *55 degrees celcius = 1702.J energy per gram actual actual temp = energy required of water per degree grams of temperature of H20
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.
k
The temperature would be that of water's boilng point od 100 degrees
Just use the definition of specific heat. Use the following formula: (energy) = (mass) x (temperature difference) x (specific heat) Replace the amounts you know, and solve for the one you don't - in this case, the specific heat.
21 grams through 71 degrees is 21x71 calories.
That completely depends on the specific heat capacity of the substance of which the sample is composed, which you haven't identified. It will take a lot more heat energy to raise the temperature of 65 grams of water 35 degrees than it would take to do the same to 65 grams of iron or yogurt, e.g.
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
No, certainly not.Temperature is a measure of the average kinetic energy of the particles in a body. The temperature of a thing is how strongly the little bits of that thing are shaking about. If they shake hard enough, meaning that the thing is hot enough, they shake the bits apart, so that the thing melts or evaporates.If I take something hot and put it against something cold, then the shaking of the molecules of the hot matter jostle the molecules of the cold matter, passing on some of their energy. To us that is a flow of heat energy from the hot matter to the cold.Get that straight! It is a flow of energy, not of temperature, and the temperature is not the flow!But, you say, suppose I take 10 grams of water at 95 degrees and put them against 10 grams of water at 35 degrees, I will get 20 grams at 65 degrees, right? How does that differ from a flow of temperature?Temperature does not flow; heat does. I chose that example carefully to make it look like a flow of temperature. Think of a different example: suppose that we put 10 grams of mercury at 95 degrees against 10 grams of water at 35 degrees; then we would get the whole lot at just about 37 degrees instead of 65 degrees, because it takes about 30 times as much heat to increase the temperature of one gram of water by one degree as it takes to heat one gram of mercury by one degree.Now, what happened to that "flow of temperature"?Get the picture?Heat will flow until the temperatures are the same all right, but the heat still is the only thing that flows.But, you say, isn't the temperature itself the flow?No, because if I have water at 95 degrees and I don't have it touching anything at a different temperature, then there is no flow of heat (or energy, if you like; same thing in our examples) and yet the temperature stays at 95. If the temperature were the flow, then zero flow would mean zero temperature, right? And do we get zero temperature? Not a bit of it; we get 95 degrees!Is this helping you get it straight? If not, ask again.
q (amt of heat) = mass * specific heat * temp. differenceThe specific heat of water is 1.00 cal/goC & the temperature difference is 70-30 = 40oCq = (105 grams)*(1.00 cal/goC)*(40oC)= 4,200 calories
q=mass * Cs * delta T = 0.25kg * 4.18 * (85-10) =78J