A 0.6-m-diameter, 1.9-m-long cylindrical tank containing liquefied natural gas (LNG) at \(-160^{\circ} \mathrm{C}\) is placed at the center of a 1.9-m-long \(1.4-\mathrm{m} \times 1.4-\mathrm{m}\) square solid bar made of an insulating material with \(k=0.0002 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). If the outer surface temperature of the bar is \(12^{\circ} \mathrm{C}\), determine the rate of heat transfer to the tank. Also, determine the LNG temperature after one month. Take the density and the specific heat of LNG to be \(425 \mathrm{~kg} / \mathrm{m}^{3}\) and \(3.475 \mathrm{~kJ} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\), respectively.

The formula for the rate of heat transfer (Q) through a cylinder is given by $$Q = \frac {2\pi kL(T_1-T_2)}{\ln \left(\frac{r_2}{r_1} \right)}$$ where L is the length of the cylinder, k is the thermal conductivity of the material, \(T_1\) and \(T_2\) are the inner and outer surface temperatures, and \(r_1\) and \(r_2\) are the inner and outer radii. Given the problem's parameters: Diameter of the cylinder: \(0.6 \,\text{m}\) Length of the cylinder: \(1.9 \,\text{m}\) Thermal conductivity: \(0.0002 \,\frac{\text{W}}{\text{m}\cdot{^{\circ}\text{C}}}\) Inner surface temperature: \(-160^{\circ} \mathrm{C}\) Outer surface temperature: \(12^{\circ} \mathrm{C}\) First, calculate the inner and outer radii. \(r_1 = \frac{0.6\,\text{m}}{2} = 0.3\,\text{m}\) (radius of LNG tank) The square solid bar is \(1.4\,\text{m}\) wide, so the insulating material thickness is \(\frac{1.4\,\text{m}-0.6\,\text{m}}{2} = 0.4\,\text{m}\). Then, \(r_2 = r_1 + 0.4\,\text{m} = 0.7\,\text{m}\). Now, substitute the given values into the formula and find Q. $$Q = \frac {2\pi (0.0002)(1.9)((12)-(-160))}{\ln \left(\frac{0.7}{0.3} \right)} = 74.88\,\text{W}$$

Find the energy gain by LNG after one month, and determine the temperature rise. The energy gain (E) can be calculated using the formula: $$E = Qt$$ where t is the time in seconds. One month has approximately \(30 \times 24 \times 60 \times 60\) seconds. Calculate the energy gain by LNG after one month: $$E = (74.88\,\text{W}) \times (30 \times 24 \times 60 \times 60\,\text{s}) = 194348160 \, \text{J}$$

To find the temperature rise, we will use the formula: $$\Delta T = \frac{E}{mc}$$ where m is the mass of the LNG, and c is the specific heat of the LNG. Given the density (\(\rho\)) and specific heat (c) of the LNG are \(425\,\frac{\text{kg}}{\text{m}^3}\) and \(3475\,\frac{\text{J}}{\text{kg}\cdot{^{\circ}\text{C}}}\), respectively. First, find the volume and mass of the LNG: $$V_\text{LNG}= \pi r_\text{LNG}^2L_\text{LNG}=\pi(0.3\,\text{m})^2(1.9\,\text{m})=0.54\,\text{m}^3$$ $$m = \rho V_\text{LNG} = (425\,\frac{\text{kg}}{\text{m}^3})(0.54\,\text{m}^3) = 229.5\,\text{kg}$$ Now, find the temperature rise of the LNG: $$\Delta T = \frac{194348160\,\text{J}}{(229.5\,\text{kg})(3475\,\frac{\text{J}}{\text{kg}\cdot{^{\circ}\text{C}}})} = 24.34^{\circ}\text{C}$$

Now that we have the temperature rise, we can calculate the final temperature of the LNG by adding the initial temperature. $$T_\text{final} = T_\text{initial} + \Delta T$$ $$T_\text{final} = -160^{\circ} \mathrm{C} + 24.34^{\circ}\text{C} = -135.66^{\circ}\text{C}$$ The rate of heat transfer to the tank is \(74.88 \,\text{W}\), and the temperature of the LNG after one month is \(-135.66^{\circ} \mathrm{C}\).