A beaker of water sits in the sun until it reaches an equilibrium temperature of 30°CThe beaker is made of 100 g of aluminum and contains 165 g of waterIn an attempt to cool this system, 100 g of ice at 0°C is added to the water.(a) Determine how much ice remains after the system reaches equilibrium.(b) Determine the final temperature of the system if instead 45 g of ice were originally added to the warm water.(in °C)
Assume both a) b) are reversable processes (isentropic and adiabatic)This allows you to consider the ice and any of it that melts independently from the water in the beaker (even though they mix)a) Assume all of the ice DOES NOT melt, therefore the system (beaker + water ice + melt) final temp T20 CFor the water initally in the beaker and the beaker to cool to 0 C, that energy has to be absorbed by the ice cube, some of which meltsNow you needs some data: Given: mw mass of water initially in beaker 0.165 kg ma mass of beaker 0.1 kg T1 inital temp of beaker system 30 C You also need some reference data: Cpw specific heat of water 4.186 KJ/kg Cpa specific heat of aluminum 0.9 KJ/kg Hsl latent heat of fusion (water) 333 KJ/kg mi mass of ice that melts (unknown) Q (leaving the water/beaker) Q (melting ice) mw Cpw (30-0) + ma Cpa (30-0) mi Hsl solve for mi: mi [mw Cpw + ma Cpa] 30 / Hsl mi [0.165 4.186 + 0.1 0.9] 30 / 333 mi 0.070 kg 70g melts therefore mass of ice remaining at equilibrium is just 100 - 70 30g Since only 70g needed to melt to absorb all of the energy lost from the 30 C beaker + water, the assumption that all the ice DOES NOT melt was a good oneb) In this part assume ALL THE ICE MELTS, therefore the final system temp of everything is in the range 30 C T2 0 CThe energy (heat) leaving the beaker + water system still has to equal the energy absorbed by the ice cube system; however, after the ice cube melts completely (at constant temp 0 C), the resulting water is then heated to T2The consevation of energy equation has one additional term: mw Cpw (30 - T2) + ma Cpa (30 - T2) mi Hsl + mi Cpw (T2 - 0) where mi 45g 0.45kg and using all the same givens data from a), everything is known except T2Substituting and skipping some math: 0.781 (30 - T2) 14.99 + 0.188 T2, solving I get T2 8.7 C Good Luck!
