A piece of Cooper rode with length of 1.5 m and radius of 5 cm heated in the one

Question

A piece of Cooper rode with length of 1.5 m and radius of 5 cm heated in the one end to 200 degreeC with total amount of energy Q=500 kJ. b) If the hot end of the rod is immersed in the mixture of the ice m_c=100 gr and 300 cm^3 water with density of 1 gr/cm^3. Determine the equilibrium temperature. Consider only 20 cm of the copper rod is immersed into the mixture of ice water and assume the temperature of the rod uniform temperature and ignore any evaporation of the water. (Density of the cooper is 8.89 g/cm^3, specific heat of copper is C_cu= 387 J/kg. degreeC, and heat conduction constant of copper is k_cu=390 w/m.degreeC and Latent heat of ice 330 kJ C_w=4.2 kJ/kg degreeC).

Explanation / Answer

(b)
Density = Mass/Volume
Mass of water = 300 * 1
Mass of water = 300 grams

Mass of copper = (20* *5^2) * 8.89 = 13964.4 grams
Let the final Temperature be Tf.

Heat Lost by copper = Heat Gained by water + Heat gained by Ice.
mc * Cc * (200 - Tf) = mw * Cw * Tf + mi * Ci * Tf + mi * Li
13964.4 * 0.387 * (200 - Tf) = 300 * 4.2 * Tf + 100 * 4.2 * Tf + 100 * 333.55
Tf = 147.86 o C
Final equilibrium temp, Tf = 147.86 o C

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