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Grade: 12th pass

                        

1g of ice at 0 degree Celsius is mixed with 1g of steam at 100 degree Celsius. After thermal equillibrium is attained the temp. Of mixture is?

3 years ago

Answers : (1)

Hamza
29 Points
							I am going to assume that both the steam and ice at at their respective temperatures to change state, ie 100C and 0C.I am also going to assume that you have knowledge oflatent heat of fusion and vaporizationspecific heat capacitySpecific heat capacity of ice, water and steam are different.And we need additional information as well. Luckily for us, we can ask Google. For water the latent heat of heat is 334J/g and the latent heat of vaporization is 2230J/g. Specific heat capacity of water is 4.2J/(gK).The steam condenses into 100C water, releasing 2230J. 334J of it goes to melting the ice into 0C water. The rest (1896J) heats up the 0C water to 1896/4.2 = 451C. But water cannot be heated up to 451C! What this means is that not all the steam needs to condense to bring the 0C ice to 100C water. And therefore the water/steam mixture is at 100C.This result is a bit counter-inituitive because most of us do not realise that 1g of steam is A LOT of steam. A quick calucation tells me it is about 1.7L. That is bigger than your soft drink PET bottle. We cannot underestimate the destructiveness of steam. It packs a punch when it condenses. 2230J/g is nothing to sneeze at.Extra, if in your version of this question, the excess heat, from condensation after melting the ice, can only bring it up to say 60C. Bear in mind that you still have 1g of 100C water from condensation. That will make 2g of 80C water.Extra. Challenge yourself. What is the mass of 100C steam required to melt 1g of 0C ice and heat it up to 100C?Heat required to melt 1g of ice = 334JHeat required to heat to 100C = 420JMass of steam required = (334+420)/ 2230 = 0.338g
						
3 years ago
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