For all problems – SHOW YOUR WORK! 1.A 500-g piece of _____________ (metal 1) is

Question

For all problems – SHOW YOUR WORK!

1.A 500-g piece of _____________ (metal 1) is heated to 120 °C and placed in an insulated vessel containing 350 mL H2O at 20 °C. Assuming no loss of water and a heat capacity of the metal from Table 6.2 in the textbook, what is the final temperature?

2.How much would the final temperature change if instead of metal 1 a chemist would use metal 2? Explain your result in context of specific heat capacities.

3.How much would the final temperature change if 250 g of metal 1 would be used instead? Explain your result.

4.When calculating a specific heat capacity, would you expect an alloy have larger or smaller specific heat capacity than water? Why?

5. A 1.00 g piece of copper metal at 150.00 °C is placed in an insulated cup containing 10.0 g of water at 18.80 °C. The final temperature of the mixture is 20.00 °C. What is the specific heat of copper? Discuss your result with respect to your answer to problem 4.

6. A) When 155 mL of water at 26°C is mixed with 75 mL of water at 85°C, what is the final temperature? (Assume that no heat is released to the surroundings; d of water = 1.00 g/mL.)

(I got 45°C).

B) Use twice the volume of each water sample. Calculate final temperature.

C) Instead of 75 mL, use 150 mL. Calculate final temperature.

D) Instead of 75 mL, use 25 mL. Calculate final temperature.

E) Discuss your results. How did the different volumes affect the final temperature?

Explanation / Answer

1.
Metal 1 is ALuminium, having C= 0.9 J/g-K
Let the final temperature be T oC
mass of water = volume * density = 350 mL* 1 g/mL = 350 g
C of water = 4.184 J/g-K

UsE:
Heat lost by metal = heat gained by water
500*0.9*(120-T) = 350*4.184* (T-20)
54000 - 450*T = 1464.4*T - 29288
T = 43.5 oC
Answer: 43.5 oC

2.
Metal 1 is Graphite, having C= 0.711 J/g-K
UsE:
Heat lost by metal = heat gained by water
500*0.711*(120-T) = 350*4.184* (T-20)
42660 - 355.5*T = 1464.4*T - 29288
T = 39.5 oC
Answer: 39.5 oC

3.
UsE:
Heat lost by metal = heat gained by water
250*0.9*(120-T) = 350*4.184* (T-20)
27000 - 225*T = 1464.4*T - 29288
T = 33.3 oC

4.
Lower heat capacity is better.
Lower the specific heat caapcity, less will be the error in measurement

we are allowed to answer Only 4 parts at a time

Related Questions

Navigate

• Browse (All)
• Browse F
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.