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15. An iron bar at 325 K is placed in a sample 20. Which sample of Fe contains p

ID: 541058 • Letter: 1

Question

15. An iron bar at 325 K is placed in a sample 20. Which sample of Fe contains particles of water. The iron bar gains energy from the water if the temperature of the water is (1) 65 K (3) 65°C (2) 45 K(4)45°C having the highest average kinetic energy? (1) 5 g at 10°C (3) 5 g at 400 K (2) 10 g at 2s C (4) 10 g at 300 K 20 21. A 36-gram sample of water has an initial temperature of 22°C. After the sample absorbs 1,200 joules of heat energy, the final temperature of the sample is 16. How much heat energy must be absorbed to completely melt 35.0 grams of H,O(s) at 0°C? (1)9.54J (3) 11,700J (2) 146] (4) 79,100J (1) 8.0°C (2) 14°C (3)30.°C (4) 55°C 6 17. What is the total number of joules released 22. What is the minimum amount of heat required to completely melt 20.0 grams of ice at its melting point? (1) 20.0J (3) 6,680 J (2) 83.6J (4) 45,200 J when a 5.00-gram sample of water changes from liquid to solid at 0°c? (1) 334 (3) 2,260J (2) 1,670] (4) 11,300J 17 23. A person with a body temperature of 37°C holds an ice cube with a temperature of 0°C in a room where the air temperature is 20.°C. The direction of heat flow is (1) from the person to the ice, only (2) from the person to the ice and air, and from the air to the ice (3) from the ice to the person, only (4) from the ice to the person and air, and from the air to the person 23

Explanation / Answer

15) iron bar gains eenrgy , if T of water is greater than 325 k

   3) 65+273.15 = 338.15 k

   4) 45+273.15 = 318.15 k

   answer: 3

16) DHfus of ice = +6.02 kj/mol

energy required = n*DHfus

                  = (35/18)*6.02*10^3

                 = 11705.55 joule

   answer: 3

17) DHfus of ice = +6.02 kj/mol

energy required = n*DHfus

                  = (5/18)*6.02*10^3

                 = 1672.2 joule

   answer: 2

20) if temperature is more kinetic eenrgy is more

   answer: 3 ( T = 400k)

21) q = m*s*DT

   1200 = 36*4.184*(x-22)

   x= 30 c

answer: 3

22) DHfus of ice = +6.02 kj/mol

energy required = n*DHfus

                  = (20/18)*6.02*10^3

                 = 6688.9 joule

answer: 3

23) answer: 2

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