1) (4 pts) The following data summarize experiments on unknown substances, code

Question

1) (4 pts) The following data summarize experiments on unknown substances, code named A, B, and C. 100.0 g of A at 75.0 °C was put in thermal contact with 100.0 g of B at 25.0 °C. When the system reached thermal equilibrium (with no heat lost to surroundings), the temperature was 35 °C. 60.0 g of B at 80.0 °C was put in thermal contact with 80.0 g of C at 30.0 °C. When the system reached urroundings), the temper 50.0 g of A at 70.0 °C is put in thermal contact with 30.0 g of C at 20.0 °C. What will be the temperature when the system reaches thermal equilibrium (with no heat lost to surroundings)?

Explanation / Answer

Let the specific heat capacities of the unknowns be SA, SB and SC.

We shall apply the principle of thermochemistry:

Heat lost by hot substance = Heat gained by cold substance

Part 1:

We have A at a higher temperature (100°C) while B is at a lower temperature (25°C). The final temperature attained is 35°C.

Heat lost by A = (mass of A)*(specific heat capacity of A)*(change in temperature of A) = (100.0 g)*SA*(75.0 – 35.0)°C = 4000*SA g.°C.

Heat gained by B = (mass of B)*(specific heat capacity of B)*(change in temperature of B) = (100.0 g)*SB*(35.0 – 25.0)°C = 100*SB g.°C.

As per the principle of thermochemistry,

4000*SA g.°C = 1000*SB g.°C

=====> 4*SA = SB …….(1)

Part 2:

We have, as per the principle of thermochemistry,

(60.0 g)*SB*(80.0 – 60.0)°C = (80.0 g)*SC*(60.0 – 30.0)°C

=====> 1200*SB = 2400*SC

=====> SB = 2*SC …….(2)

Comparing (1) and /92), we have,

4SA = SB = 2SC

=====> 4SA = 2SC

=====> 2SA = SC ……..(3)

Let the temperature attained be t°C. Apply the principle of thermochemistry to obtain

(50.0 g)*SA*(70.0 – t)°C = (30.0 g)*SC*(t – 20.0)°C

====> 50*SA*(70.0 – t) = 30*SC*(t – 20.0)

====> 50*SA*(70.0 – t) = 30*2*SA*(t – 20.0)

====> 50SA*(70.0 – t) = 60*SA*(t – 20.0)

Since SA is not equal to zero, we have,

50*(70.0 – t) = 60*(t – 20.0)

====> 3500 – 50t = 60t – 1200

====> 3500 + 1200 = 60t + 50t

====> 4700 = 110t

====> t = 4700/110 = 42.727 42.73

The temperature attained is 42.73°C (ans).

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