A heavy copper pot of mass 2.0 kg (including the copper lid) is at a temperature of 150°C, you 0.10 kg of water at 25°C into the pot, then quickly close the lid of the pot so that no steam can esca Find the final temperature of the pot and its contents, and determine the phase (liquid or gas) of the water. Assume that no heat is lost to the surroundings.
Answers
Answer:A heavy copper pot of mass 2.0 kg (including the copper lid) is at a temperature of 150°C, you 0.10 kg of water at 25°C into the pot, then quickly close the lid of the pot so that no steam can esca Find the final temperature of the pot and its contents, and determine the phase (liquid or gas) of the water. Assume that no heat is lost to the surroundings.
Explanation:
Answer:
this question we have a heart corporate part of mass m equals 2.0 Kg including the lid. And it is their deep temperature. Oh 1 50° integrate. Okay Now you pour the mass of water equals to 0.10 kg of old water, Add the water. It is equals to 25° integrate onto the board. Then quickly replaced that lives who knows, team can escape. So we have to determine the final temperature of the sport and it's contained and also the face. Okay, so we can write from here that the equilibrium temperature. Okay so for the calculation of the equilibrium temperature we can right that the final will be a question. The mm what a place to be specific heat of the copper but player B initial temperature Okay and this is less mass of water manipulated the specific heat of water, manipulated temperature the the blue. Okay and develop a so uh let us do this into the another way. Okay, so first of all we can find that the heat lost by the corpus. This is equals to amber. Playerbase specific heat of the copper. Okay, that completes the difference. Okay, they're 30. So uh Let us suppose that temperature has reached two final temperature equals 200° integrated. Okay, this has been supposed Okay so mm which is 2.0 specific heat of the copper which is three double seven joules per program degrees integrate and this is 100. Okay and this is from the temperature 1 15 so we can write that this is 1 15 minus country. So the released heat from here or the last heat from here will be equals to pave double seven double zero through. Now the heat gained by this water. So heat gained will be equals two long 15 C to first of all the 100 C. Okay so M. W. C. W. And this is the final minus D. W. Okay, so M W is 0.1 kg. C. W. Is 4186. And the final 800 -25. So he gained from here comes out to be 31395 june Okay so this is the question. So it means the last hit is this much and the game heat is this much. Okay so we can find that the difference of the heat that is the remaining heat. Okay, you remaining? Okay this will be equal to the You lost mine rescue game. So this is three double 7 double seven double zero minus 27395. So this comes out to be 1605 ju which means this is the remaining heats which means temperature of 100 degrees integrated will achieve. And this heat will convert convert water into steam Okay Into steam. But for the conversion of this 0.1 Kg the heat required will be 0.1 but player baby 26 zero particular we turned to the power three so it means this is 26,000. So this is very high quantity than this, which means full Whole water will not be converted. The final temperature will be equals 200° integrated. And the face is what? Less speed. Okay, So this is the answer for the problem here. Okay. Thank you.