3) Text (17.82) You pour 108 cm3 of ethanol, at a temperature of -10 o C, into a

Question

3) Text (17.82) You pour 108 cm3 of ethanol, at a temperature of -10 o C, into a graduated cylinder initially at 20 o C, filling it to the very top. The cylinder is made of glass with a specific heat 0.84 kJ/kg.K and a coefficient of volume expansion of 1.2 X 10-5 / K. its mass is 0.110 kg. The mass of the ethanol is 0.0873 kg. (a) What will be the final temperature of the ethanol, once thermal equilibrium is reached? (b) How much ethanol will overflow the cylinder before thermal equilibrium is reached? (From Table 17.2 and 17.3, ethanol volume expansion coefficient 75 X 10-5 / K, ethanol specific heat 2428 J/kg.K ).

4) As shown in figure below, suppose that the rod in the figure is made of Cu, if 45.0 cm long and has a cross-section area of 1.25 cm2 . Let TH = 100 o C and TC = 0.0 o C. (a) What is the final steady-state temperature gradient along the rod? (b) What is the heat current in the rod in the final steady state? (c) What is the final steady-state temperature at a point in the rod 12.0 cm from its left end?

5) Text (17.101) A copper calorimeter can with mass 0.446 kg contains 0.0950 kg of ice. The system is initially at 0.0 o C (a) if 0.0350 kg of steam at 100.0 o C and 1.00 atm pressure is added to the can, what is the final temperature of the calorimeter can and its contents? (b) At the final temperature, how many kilograms are there of ice, how many of liquid water, and how many of steam? (From Table 17.3 and 17.4, water latent heat of vaporization 2.256 X 106 J/kg, water latent heat of fusion 3.34 X 105 J/kg, water specific heat 4.19 kJ/kg-K, copper specific heat 390 J/kg-K).

Explanation / Answer

Equate heat exchange by ethanol & cylinder using the relation ( Heat transferred = mass x specific heat x drop in temperature), then we get-

0.0873 x 2428 x (Tf - 263 ) = 0 .110 x 0.84 (293 -Tf)

Tf = 263.02 K

or, Tf = - 9.97^o C   

Where Tf represents the final temperature.

Now,find out the volume change in cylinder as well as ethanol separately due to variation in temperature , calculate as above

Total volume of cylinder at Tf (ie. 263 02 K) = 108 x1.2 x 10^-5 x 263 02 = 0.34087 cm 3

total volume change of ethanol at Tf = 108 x75 x 10^-5 x 263.02 = 2.130462 cm ^3

So, volume of ethanol which would overflow due to variation in temperature = 2.130462 - (-.34087 )

------------------------------------------------------------------------------------------------ = 2.47 cm^3

Note that there would be an expansion in volume of ethanol , whil the volume of cylinder would contract,hence a negative volume change is considered.

* Glad to help , but please post the other two questions separately as new questions.

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