Enthalpy change

Scarbo

2011-10-14 6:17 am

1. 69.3 kJ heat were added to a0.500kg sample of aluminium at 27.0℃.What would be thefinal temperature of the sample?(Specific heat capacity of aluminiun=0.900J g^-1 K^-1)A. 15.4℃B. 42.4℃C. 154℃D. 181℃ 2. The specific heat capacities ofsome metals are shown below.Aluminium: 0.900J g^-1 K^-1Gold: 0.13 J g^-1 K^-1Copper: 0.390 J g^-1 K^-1Mercury: 0.140 J g^-1 K^-1 If the same amount of heat were applied to the same mass of eachmetal, which metal would show the smallest temperature change? ANS: AA. AlB. Au C. CuD. Hg 3. A gold sample of 84.1g at 120℃ is placed in 106g of water at 21.4℃. What isthe final temperature of the system? ANS:D(Specific heat capacities: gold =0.130 J g^-1 K^-1, water=4.18 J g^-1K^-1)A. 70.8℃B. 65.0℃C. 27.3℃D. 23.8℃ 4. A 160g sample of copper at some temperature was added to 50.0 g ofwater. The initial water temperature was 25.0℃ and thefinal temperature was 30.5 ℃. What was theinitial temperature of the copper sample? ANS:D(Specific heat capacities: copper=0.390 J g^-1 K^-1, water=4.18 J g^-1K^-1)A. 17.6℃B. 18.4℃C. 48.1℃D. 48.9℃5. A 23.0g sample of sliver washeated to 98.8℃ in boiling water and then dropped into abeaker containing 45.0g of water at 25.0 ℃ and the finaltemperature was 27.1℃. ANS:C(Specific heat capacity of water=4.18 J g^-1 K^-1) A. 0.122 J g^-1 K^-1B. 0.137 J g^-1 K^-1C. 0.240 J g^-1 K^-1D. 2.79 J g^-1 K^-1How can I get the answers?
Thanks!!

回答 (1)

用戶26240

2011-10-14 7:00 am

✔ 最佳答案

1.
The answer is : D. 181°C

Let T°Cbe the final temperature.
ΔE = m c ΔT
69.3 ´1000 = (0.5 ´ 1000)´0.9 ´ (T- 27)
T = 181
Final temperature = 181°C

= = = = =
2.
The answer is : A. Al

ΔE = m c ΔT
ΔT = (ΔE/m) (1/c)

For ΔE and m are constant, ΔT = constant/c
Hence, ΔT ∝1/c
As aluminum has the greatest c, thus aluminium shows the smallest temperaturechange.

= = = = =
3.
The answer is : D. 23.8°C

Let T°Cbe the final temperature.

Heat lost by gold = Heat gained by water
84.1 ´0.130 ´(120 - T) = 106 ´4.18 ´ (T- 21.4)
1311.96 - 10.933T = 443.08T - 9481.912
454.013T = 10793.872
T = 23.8
Final temperature = 23.8°C

= = = = =
4.
The answer is : D. 48.9°C

Let T be the initial temperature of the copper sample.

Heat lost by copper = Heat gained by water
160 ´ 0.390 ´ (T - 30.5) = 50.0 ´ 4.18 ´ (30.5 - 25.0)
62.4T - 1903.2 = 1149.5
62.4T = 3052.7
T = 48.9
Initial temperature of the copper sample = 48.9°C

= = = = =
5.
The answer is : C. 0.240 J g⁻¹ K⁻¹

Let c be the specific heat capacity of silver.

Heat lost by copper = Heat gained by water
23.0 ´ c ´(98.8 - 27.1) = 45.0 ´ 4.18 ´ (27.1 - 25)
C = 0.240
Specific heat capacity of water = 0.240 J g⁻¹ K⁻¹

參考: 土扁

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