Thank you so much! I really like how that approach allowed me to check what process was happening based on available energy. I was having difficulty thinking of loss of energy from the water as 'available energy' for the ice because I always thought of mc ΔT as heat loss, but for the ice, this...
Using that correct calculation, to both change the ice to 0°C and melt it all would be:
(100(2.1)(15) + (100)(334) = 36550 J
This would correspond to a heat loss in the water:
36550 = (200)(4.2)(ΔT)
change in T = 43.5 K
This is too large a ΔT; the water would drop below freezing and even be...
Hi, I actually have the same problem so I'm going to try at items 2 and 3 ;)
2. Q = mcT and Q=mL
3. I reasoned that heat flows from the water to the ice cube, and the heat required to get the ice cubes to 0°C (used specific heat of water) and to melt them to all liquid is
(100g)(2.1J/gK)(15 K)...