A-level physics thermodynamics/ heat capacity/ latent heat difficult problem

Specific heat capacity of water: 4180 J/kg/KSpecific heat capacity of ice: 2030 J/kg/KSpecific latent heat of fusion of ice: 3.35×10^5 J/kg/KSpecific latent heat of vaporization of water: 2.26×10^6 J/kg/KA mass of 0.35kg of ice at −15°C is lowered into an insulated beaker containing a certain mass (in kg) of water at 59°C.What is the minimum mass of water at 59°C needed in the beaker to achieve a final temperature of 0.0°C? What is the maximum mass?My answer (which is wrong):There are 3 possible equations, 1 containing LΔm for both ice and water, one containing LΔm for just ice, and one containing LΔm for just waterEquation 1: mcΔT (ice) LΔm (ice) = mcΔT (water) LΔm (water) = m(cΔT L) (water)Equation 2: mcΔT (ice) LΔm (ice) = mcΔT (water)Equation 3: mcΔT (ice) = mcΔT (water) LΔm (water) = m(cΔT L) (water)The minimum mass of water is when the LHS is a minimum and the RHS is a maximum, i.e equation 3 :frown:

0.35) (2030) (15) = m (4180 x 59 2.26 x 10^6) --> solving for m, m = 4.25 x 10^-3The maximum mass of water is when the LHS is a maximum and the RHS is a minimum, i.e equation 2 :frown:

0.35) (2030) (15) (3.35x10^5) (0.35) = m x (4180) (59) --> solving for m, m = 1.40 kgBoth of my answers are wrong, but I don't understand why.Thank you!

Reply 1

Joinedup

For the minimum question maybe think about flipping it and gradually dripping the hot water into a beaker of crushed ice, the liquid water would cool and then start to freeze, the liquid water freezing would put some heat into warming up the original ice wouldnt it? If you kept that up for a while you'd end up with all ice at zero celcius.

The question is a bit sneaky cos as its described I'd sort of imagined convection currents stirring the liquid water around a big lump of ice in a way that would always be bringing new hot water into contact with the original ice.