heat to melt ice at zero C = 850 x heat fusion = ?
heat removed from 35 C water = 1,700 x specific heat water x (0-35) = ?
I get about 280,000 J to melt the ice and about 250,000 J from the water. The ice won't melt completely. The end solution will be a mixture of ice and water at zero C. You can calculate how much ice is left if you wish. There is enough data to do that.
An 850 gram chunk of ice at 0 degrees Celsius is dropped into a container holding 1.7 kg of water at an initial temperature of 35 degrees Celsius. Due to the presence of the ice, the temperature of the water eventually drops to 0 degrees Celsius. Show a mathematical solution to answer this question:
Does the 850 gram chunk of ice completely melt?
Formulas: Q= mHf Q=mHv Q= m∆TCp
The specific heat of liquid is 4.18J/gᵒC
The specific heat of solid water (ice) is 2.11J/gᵒC
The heat energy released during melting is 334J/g
3 answers
Could you please prove how the ice doesn't completely melt? Don't you subtract 280,000 J and 250,000 J and get 30,000 J? and then you divide it by 334 J/g and get 105 g? But I don't know what you do next.
I caution you not to use my estimates. Those are just close numbers; you need to run them yourself.
What I did above SHOWS that the ice doesn't completely melt. Yes, you subtract 280,000 - 250,000 and show that you are 30,000 J short of having enough heat to melt all of the ice. I don't think anything else is necessary. Again, that 30,000 is an estimate. You need to run that number too.
What I did above SHOWS that the ice doesn't completely melt. Yes, you subtract 280,000 - 250,000 and show that you are 30,000 J short of having enough heat to melt all of the ice. I don't think anything else is necessary. Again, that 30,000 is an estimate. You need to run that number too.