Question

1.A thin rod consists of two parts joined together. One-third of it is silver and two-thirds is gold. The temperature decreases by 24.7 C°. Determine the fractional decrease ?L/(LSilver + LGold) in the rod's length, where LSilver and LGold are the initial lengths of the silver and gold rods.

Δ LGold L Gold Heat flow Heat flow L Silver Δ1-silver -13. T -Silver

2.Equal masses of two different liquids have the same temperature of 20.5 °C. Liquid A has a freezing point of -68.5 °C and a specific heat capacity of 1840 J/(kg C°). Liquid B has a freezing point of -81.9 °C and a specific heat capacity of 2670 J/(kg C°). The same amount of heat must be removed from each liquid in order to freeze it into a solid at its respective freezing point. Determine the difference Lf,A - Lf,B between the latent heats of fusion for these liquids.

3. The column of mercury in a barometer (see Figure below) has a height of 0.760 m when the pressure is one atmosphere and the temperature is 0.0 °C. Ignoring any change in the glass containing the mercury, what will be the height of the mercury column for the same one atmosphere of pressure when the temperature rises to 30.5 °C on a hot day? Hint: The pressure in the barometer is Pressure = ?gh, and the density ? changes when the temperature changes.

Answer 1

Given theta_1 = theta_2 material implication m S_A Delta theta_A + m L_A = m S_B Delta theta_B + mL_B material implication 1840 (20.5 - (-68.5)) + L_A = 2670(20.5 - (-81.9)) material implication 163760 + L_A = 273408 + L_B L_A - L_B = 109648 J/kg Given rho, g(0.76) = 1 atm When temp becomes 30.5 degree density charges Now, p = rho_30.5 g h^1 = 1 atm Thus, rho_30.5 h^1 = rho middot g(0.76) rho_0/1 + y Delta T h^1 = rho_0(0.76) h^-1 = 0.76 (1 + y(Delta T)) = 0.76 (1 + 0.00018 times 30.5) = 0.7641 m Note: Coefficient of vol expansion (y) for mercury has been taken from engineering tool box. com \r\n L_t = L_0 (1 + alpha Delta theta) Therefore L_GT = L_G_0 (1 + 14 times 10^-6 times 24.7) L_GT/L_G_0 = (1 + 0.000345) L_G_t/L_G_0 - 1 = 0.000345 delta L_G/L_G = 0.000345 rightarrow (1) Similarly, L_S_t = L_S_0 (1 + 18 times 10^-6 times 24.7) L_S_t/L_S_0 - 1 = 0.000444 delta L_S/L_S = 0.000444 rightarrow (2) From (1) and (2) delta L_G + delta L_S = 0.000345 L_G + 0.000444 L_S = 0.000345 times 2/5 L + 0.000444 L/S delta L/L = 0.000378