If l have 10.0 g of isopropranol at -100.0°C, how much energy is required to rai

Question

If l have 10.0 g of isopropranol at -100.0°C, how much energy is required to raise the temperature to 60.0°C. Note: determine the freezing and boiling points from the graph.

Name: ID:B Use the following heating curve for isopropanaol to answer the next two questions. Note that solid, liquid, and gaseous ethanol have different specific heat values. Cpup = 0.410 cal gC IL-41.76 cal Temperature H_=21.51 cal CPpa _0.7 15 cal g"C CPolid = 0.309 cal SC Time 22. (3 points) Please choose the correct phases of isopropanaol. Point B: (solid) (liquid) (gas) Point C: (solid) (liquid) (gas) Point F: (solid) (liquid) (gas) 23. (8 points) If I have 10.0 g of isoprogramacol at 100.0°C, how much energy is required to raise the temperature to 60.0°C. Note: determine the freezing and boiling points from the graph.

Explanation / Answer

The freezing and the boiling points of isopropanol can be determined from the graph as -88.5°C and 82.6°C respectively. The overall heat required is the sum of the heat required for the following processes.

i) Heat required to raise the temperature of solid isopropanol from -100°C to -88.5°C, q1 = (mass of isopropanol)*Cp(solid)*(change in temperature) = (10.0 g)*(0.309 cal/g.°C)*[(-88.5°C) – (-100°C)] = 35.535 cal.

ii) Heat required to melt the solid isopropanol at -88.5°C, q2 = (mass of isopropanol)*Hfus = (10.0 g)*(21.51 cal/g) = 215.1 cal.

This is the heat of fusion of solid isopropanol.

iii) Heat required to raise the temperature of liquid isopropanol from -88.5°C to 60°C, q3 = (mass of isopropanol)*Cp(liquid)*(change in temperature) = (10.0 g)*(0.715 cal/g.°C)*(60°C – (-88.5°C)] = 1061.775 cal.

The total heat required is q = q1 + q2 + q3 = (35.535 + 215.1 + 1061.775) cal = 1312.41 cal (ans).

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