Physics Heat (Change of state)

刀光劍影

2009-06-07 6:37 pm

An experiment was set up to demonstrate the heat transfer between two objects in contact. At the end of experiment, it was found that 1 kg copper block and 100 g water was found in a beaker.The temperature of water was found at 3 °C and did not change anymore.

(given specific heat capacity of copper=385Jkg–1 K–1, water=4200Jkg–1 K–1, ice=2100Jkg–1 K–1,specific latent heat of fusion of water=0.334 kJ g–1)

Q.If the temperature of copper block was 420 K at the beginning of the experiment , find the temperature of H2O(liquid state is water) at the beginning of the experiment.

P.S. Please give the full solution!!

回答 (1)

Prof. Physics

2009-06-07 10:02 pm

✔ 最佳答案

In the view of conservation of energy,

Heat loss by the copper = Heat gain by H2O

Now, heat loss by the copper

= mcT

= (1)(385)[(420 - 273) - 3]

= 55 440 J

If the water is at 0*C initially, then the heat gain by the water

= mcT

= (0.1)(4200)(3 - 0)

= 1260 J

Now, if the H2O is ice at 0*C initially, then the heat gain

= ml + mcT

= (0.1)(3.34 X 105) + 1260

= 34 660 J

It is still less than the heat loss by the copper.

So, the H2O should be ice below 0*C initially.

And mcT + 34 660 = 55 440

(0.1)(2100)(0 - T) + 34 660 = 55 440

T = -99*C

Therefore, the initial temperature of the H2O is -99*C, which is in solid state (ice.)

參考: Physics king

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