A.) A 4.1-kg block of ice originally at 263 K is placed in thermal contact with

Question

A.) A 4.1-kg block of ice originally at 263 K is placed in thermal contact with a 13.1-kg block of silver (cAg = 233 J/kg-K) which is initially at 1064 K. The H2O - silver system is insulated so no heat flows into or out of it.

-At what temperature will the system achieve equilibrium?

_____K

-What will be the phase of the H2O at equilibrium?

Solid (ice)

Liquid (water)

Gas (vapor)

B.) An ideal gas in a piston is placed in a refrigerator, which removes 526 J of energy from the gas via cooling. The plunger on the piston is pushed down, which does 122 J of work on the gas.

-What is the net change in the internal energy of the gas in the piston?

______J

Explanation / Answer

A. Heat lost by Ag = Heat gained by Ice

mcDT = mL + mcDt

13.1 * 233 * (1064 - Tf) = (4.1 * 3.33e 5) + (4.1 *4186* (Tf-263)

3.24 e 6   - 3052 Tf = 1.365 e 6    + 17162.6 Tf   - 4.513 e 6

(17162.6 + 3052)Tf = 1.365 + 4.513 + 3.24

Tf = 9.118 e 6/(20214.6)

Tf = 451.05 deg K

------------------------------

gaseous state   as Temp T = 451-273 = 178 deg c
---------------------------

B. Use dQ = dU + dW

so

dU = 526-122

dU = 404 Joules
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