Question

1. A styrofoam cooler is a perfect cube (i.c. normal box shape) that is 0.35 m on as side. The thickness of the sides is 0.0055m. Assume the cooler also has a top (or lid) of the same thickness as the sides. The cooler is made completely of styrofoam (k = 0.010 J/sm."C). The cooler is partially filled with 15,0 liters of water (m- 15.0 kg) and 5.00 kg of ice. a. What is the total surface area of the cooler? b. If the outside temperature is 29.0°C, and the temperature inside the cooler is 7.00°C, then how much heat will the inside contents of the cooler absorb through all of its sides in one hour? c. If the amount of heat found in part b is absorbed only by the ice, then how much of the ice will melt? d. As stated above, the temperature of the water inside the cooler is initially 7.00°C. What will the final temperature of the water be after the amount of ice found in part c melts? (Ignore the rest of the ice in this calculation. Assume that the melted ice is now water that begins at 0.00°C.) (important statistics for water: p= 1000 = 1.00 4: c= 1.00 = 4186 c; L=333 L.-2260

Answer 1

@ Surface Area of calbe (cooler) side length = 0.35m Surfere area of one face & A= (0.35m) - 0.1225m² Total surface area of Coller = GA (AT) - 68011225m² =0.735m² =0.7359 0.74m² I Total surface Ares, Heat flow Rate of where k Al Pontsider k=0.010 mic I = k Apl Coutside - Oinnite) -60-010) (0.74) (302 - 280) 0.0055 A 0.35m2 Az = 0.74me e = 0.005sm Oduts to = 29+273 = 302R Oimido =2+273 olt ♡ . 29469 Ise as t= Ihr = 36oosee Q = (29.6) t 280k 8 = (2916) (3600) Joule = 106, = 106560 Joule

o ef is let amount of Heat abserhed in part b absorb by only Ice m te the mass of ice melt thon &- Q = mit 4= 333 kJ/kg 8=106560 J 106560 = m (333X103) m= 106560 3338103 m=320x10 kg mais of ice melt. m= 0+320kg 109 θε ο meeted ice mals of wath M=15.0kg .* Rest un melted mais Negle et frital temp. of water = 76 mais of ice meet into water = -0.320 kg Temp of melted ice oc let to be teme final temperature of monture then & Release by water = Sabsorbed by water Joomed by M4 (7-t) = m & (t-o) 1517-t) = 0.320t 165 lista ozet 105-15 t=0.3204 105=(15.32o) t t = 105 15.320 t = 6.85°C