1. An aluminum cube of mass 4425 g, at 22.3°C. is immersed in a Styrofoam cup fi
ID: 3279795 • Letter: 1
Question
1. An aluminum cube of mass 4425 g, at 22.3°C. is immersed in a Styrofoam cup filled with LN2 at -195.8° C. After 120 seconds (sec), 34.50 g of LN2 evaporates due to heat transferred from the cube. Calculate Lv of LNz (in units of calories and grams) and the rate R (calories/second) at which heat from the cube evaporates LN2. (Note that the 34.50 g evaporated is due solely to the immersion of the cube, and not by heat transferring into the cup from the room.) Use the accepted value of cAI (for this temperature range) when calculating L. Use your calculated value of Ly when calculating the rate R. Show formulas and calculations Show formulas and calculations (Here, R is the rate of evaporation.) (Both answers should have 2 numbers after the decimal point.) An aluminum cube of mass 44.25 g is immersed in liquid Nitrogen and comes to thermal equilibrium with the liquid Nitrogen. This equilibrium temperature is that of liquid Nitrogen, -195.8° C. The cube is then immersed in 189.47 g of water at 22.3° C. The final, equilibrium temperature of the water-aluminum cube system is 14.0° C. Calculate the specific heat of the aluminum cube. Answer should be in units of calories, grams and °C. Show formulas and calculations. 2.Explanation / Answer
Heat supplied by aluminium:
nCvdT=(44.25/26.99)*24.4*(22.3-(-195.8))=8724.81 J
Latent heat of nitrogen = Q/m = 8724.81/34.5 = 252.89 J/gm=1062.15 cal/gm
Rate of evaporation = (8724.81*4.2/120)= 305.37 cal/sec
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