Look up the heat of fusion of ice. I think it is about 335 Joules/gram but you need the exact number. It should be in a table in your text or you can find it on the web. If a cup is about 236 mL volume (again, look it up) and it is half filled with ice, that would be about 118 mL ice or 118 grams depending upon the ice being shaved, crushed, cubes, etc and how it is packed in the cup.
The energy required to melt the ice is
q = mass x heat fusion ice = approximately 118 grams x 335 Joules/gram = ?? Joules.
Now a watt is 1 Joule/sec. How many seconds will be required to furnish ?? Joules of energy to melt the ice. I know there are a number of approximations here but your question is not very specific. Post your work if you get stuck. If you have more specific information you will want to use those values in the calculation.
If a cup is half full of ice, how long will it take to melt the ice in a microwave oven (1000 W) set on high?
Also, if you did the above but added 100ml of ice cold water to the ice would this change the results of melt time?
2 answers
Microwaves that say 1000w mean input power, not microwave output power.
OK, assume some mass of water, say 300g ice at -6C
Figure the time to melt.
300*2(6)+300*334=1000*timeinseconds
I estimated the specific heat of ice at 2j/g, I don't remember it exactly.
so figure time in seconds from that.
Now, do it again, with with ice water (0C)
300*2(6)+300*334+100*c*0=1000*timeinseconds
Hmmm. Notice the ice water does not enter into the equation, as it does not heat up.
OK, assume some mass of water, say 300g ice at -6C
Figure the time to melt.
300*2(6)+300*334=1000*timeinseconds
I estimated the specific heat of ice at 2j/g, I don't remember it exactly.
so figure time in seconds from that.
Now, do it again, with with ice water (0C)
300*2(6)+300*334+100*c*0=1000*timeinseconds
Hmmm. Notice the ice water does not enter into the equation, as it does not heat up.