Use the worked example above to help you solve this problem. Suppose 0.410 kg of

Question

Use the worked example above to help you solve this problem. Suppose 0.410 kg of water initially at 35.0°C is poured into a 0.300 kg glass beaker having a temperature of 25.0°C. A 0.500 kg block of aluminum at 37.0°C is placed in the water, and the system insulated. Calculate the final equilibrium temperature of the system.

A 15.5 kg gold bar at 28.0°C is placed in a large, insulated 0.800 kg glass container at 15.0°C with 2.00 kg of water at 25.0°C. Calculate the final equilibrium temperature.

Please answer "Practice It" section and "Exercise" section! Thank you so much!

i safari File Edit View History Bookmarks Window Help Q 54% ?. Tue 4:00:25 PM Q.OE webassign.net Phy 1010 -Ch 11 Hom-W18 l Safari File Edit View EXAMPLE 11.4 Calculate an Equilibrium Temperature dmg Anki GOAL Solve a calorimetry problem involving three substances at three different temperatures PROBLEM Suppose 0.400 kg of water initially at 40.0°C is poured into a 0.300-kg glass beaker having a temperature of 25.0°C. A 0.500-kg block of aluminum at 37.0°C is placed in the water and the system insulated. Calculate the final equilibrium temperature of the system. STRATEGY The energy transfer for the water, aluminum, and glass will be designated Qws Qal and 2, respectively. The sum of these transfers must equal zero, by conservation of energy. Construct a table, assemble the three terms from the given data, and solve for the final equilibrium temperature, T. Screen Shot 018-03..6.13 PM SOLUTION Apply the final temperature equation to the system Screen Shot 2018-03..6.06 PM Construct a data table Q (J) m (kg) 0.400 c (J/kg·°C) 4190 T; T 40.0°0 l0.500 9.00 x 102 T 37.0°C Screen Shot 2018-03. 6.16 PM 0.300 837 T 25.0°C Using the table, substitute into Equation (2) (1.68 J 105 J/oC)(T 40.0°C) (4.50 x 102 J/ec)(T - 37.0°c) (2.51 x 102 J/oc)(T - 25.0°c)o (1.68 J x 1034.50 x 102 J/o(c + 2.51 x 102 J/oc)T9.01 x 104 ] T= 37.9°C 27

Explanation / Answer

Practice it:
given

Cw = 4190 J/(kg C)

Cal = 900 J/(kg C)

Cg = 837 J/(kg C)

Mw = 0.41 kg

Mal = 0.5 kg

Mg = 0.3 kg

let T is the final equilibrum temperature.

Heat lost by aluminum + heat lost by water = heat gained by glass

M_al*Cal*(37 - T) + Mw*Cx*(35 - T) = Mg*Cg*(T - 25)

0.5*900*(37 - T) + 0.41*4190*(35 - T) = 0.3*837*(T - 25)

==> T = 34.3 degrees celsius

Excercise :

given

Cw = 4190 J/(kg C)

Cgold = 129 J/(kg C)

Cg = 837 J/(kg C)

Mw = 2 kg

Mgold = 15.5 kg

Mg = 0.8 kg

let T is the final equilibrum temperature.

Heat lost by gold + heat lost by water = heat gained by glass

M_gold*Cgold*(28 - T) + Mw*Cx*(25 - T) = Mg*Cg*(T - 15)

15.5*129*(28 - T) + 2*4190*(25 - T) = 0.8*837*(T - 15)

==> T = 24.9 degrees celsius

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