The key to solving this problem is that the heat lost by the warmer water is equal to the heat gained by the colder water. Knowing the relation Q = mc(change in T), this problem can be solved with algebra. Q is heat, m is mass in grams, c is the specific heat of water 4.184 J/ g C, and T is temperature. 100g * 4.184 J / g C * (x - 30 C) = 50 g * 4.184 J / g C * (60 C - x). The final answer should have x = to 40 degrees C.
A two q's to 0 problem
q = mass*specifice heat*change in temp.
(100 g H2O)(4.180 J/gC)(Tf - 30 C) + (50 g H2O)(4.180 J/gC)(Tf - 60 C) = 0
distribute
(418Tf - 12540) + (209Tf - 12540) = 0
drop parentheses and add all together
527Tf - 25080 = 0
add 25080 to each side
527Tf = 25080
divide both sides by 527
Temperature final = 48 degrees Celsius
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How many joules of energy is needed to raise the temperature 20 grams of iron form 20 c to 50c
q(joules) = mass * specific heat * change in temperature
q = (20 grams)(4.180 J/gCo)(30o C - 25o C)
= 418 joules of absorbed energy
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Q=m.c.(Temperature 2-Temperature 1)
Specific heat capacity of water = 4200 J/ Kg . o C
= 4.2 J/ g. o C
Q= 20 g * 4.2 J/ g. o C * 5
Q= 420 J
20degrees celsius
35
31.96
2.5
It is not possible to answer the question without information about the specific heat of the metal and of the cup.
I'll assume here that by "70 temperature" you mean "70 degrees Celsius". Basically, you have to calculate the average temperature of all of the water in the mixture, which will be the final temperature once it's well stirred. The 200 grams of water at 10 degrees represent 2/3 of the total amount of water (300 grams), so thus, multiply 10 by 2/3 to determine their contribution to the final temperature. You will get 20/3. The 100 grams of water at 70 degrees represent 1/3 of the total amount of water, so multiply 70 by 1/3 to determine their contribution to the final temperature. You will get 70/3. When you add together the two temperatures you get 90/3, which is equal to 30. Therefore, the final temperature is 30 degrees Celsius.
Liters measure volume. Grams are a measure of mass, degrees Celsius are a measure of temperature, and meters are a measure of length.
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.
1000
The temperature would be that of water's boilng point od 100 degrees
It is not possible to answer the question without information about the specific heat of the metal and of the cup.
I'll assume here that by "70 temperature" you mean "70 degrees Celsius". Basically, you have to calculate the average temperature of all of the water in the mixture, which will be the final temperature once it's well stirred. The 200 grams of water at 10 degrees represent 2/3 of the total amount of water (300 grams), so thus, multiply 10 by 2/3 to determine their contribution to the final temperature. You will get 20/3. The 100 grams of water at 70 degrees represent 1/3 of the total amount of water, so multiply 70 by 1/3 to determine their contribution to the final temperature. You will get 70/3. When you add together the two temperatures you get 90/3, which is equal to 30. Therefore, the final temperature is 30 degrees Celsius.
I believe it will be 145.52 degrees Celsius if I did my math correctly. You need to convert calories to joules. I believe one joule raises the temp of 1 gram water by 1 degree Celsius so 1200*4.184=5020.8 J /40grams=125.52 temp increase+20=145.52 degrees Celsius.
If coffee and milk have the same thermal coefficient then: (15*22 + 185*86)/200 = 81 (81,2 rounded off due to significant digits)
A calorie is the amount of heat you need to raise the temperature of one gram of water by one degree Celsius. Assuming you are raising the temperature of the water from twenty degrees Celsius to ninety-nine degrees Celsius, it would take 20,000 calories. To calculate this, subtract 20 from 99. This is the amount of degrees you need to raise the temperature of the water by. Then multiply that number by 256, the amount of water in grams. You should get 20,244 calories. In significant digits, your answer should be 20,000 calories.
Do some converting first. 688 calories (4.184 Joules/1 calorie) = 2878.592 Joules 25 ml of water = 25 grams q(Jolules) = mass * specific heat * (Temp. final - Temp. initial) 2878.592 Joules = 25 grams Water * 4.180 J/gC * (Temp Final - 80C ) 2878.592 Joiles = 104.5( Temp. Final) - 8360 11238.592 =104.5(Temp. Final) 107.55 Celsius Final Temperature ( call it 108 C )
200 grams
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
105C
Approx 4974 Joules.
Liters measure volume. Grams are a measure of mass, degrees Celsius are a measure of temperature, and meters are a measure of length.