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The 35-cm-thick roof of a large room made of concrete $\left(k=0.79 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \alpha=5.88 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}\right)$ is initially at a uniform temperature of \(15^{\circ} \mathrm{C}\). After a heavy snowstorm, the outer surface of the roof remains covered with snow at \(-5^{\circ} \mathrm{C}\). The roof temperature at \(12 \mathrm{~cm}\) distance from the outer surface after $2 \mathrm{~h}$ is (a) \(13^{\circ} \mathrm{C}\) (b) \(11^{\circ} \mathrm{C}\) (c) \(7^{\circ} \mathrm{C}\) (d) \(3^{\circ} \mathrm{C}\) (e) \(-5^{\circ} \mathrm{C}\)

Short Answer

Expert verified
Answer: (d) 3 °C

Step by step solution

01

1. Convert all units to SI system

To solve the problem, we need to convert all units to the SI system before we start the calculations. Thickness of the roof = 35 cm = 0.35 m Distance from the outer surface = 12 cm = 0.12 m Temperature Initial = 15°C = 288 K Temperature outer surface = -5°C = 268 K Time = 2 hours = 7200 seconds
02

2. Heat diffusion equation

To find the temperature T(x,t), we will use the heat diffusion equation: T(x,t) = Ti + (T0 - Ti)(1 - erf(x/(2*√(α*t)))) where Ti = initial uniform temperature (288 K) T0 = outer surface temperature (268 K) α = thermal diffusivity of concrete (5.88 × 10^(-7) m^2/s) erf is the error function.
03

3. Calculate erf(x/(2*√(α*t)))

First, we calculate x/(2*√(α*t)) = 0.12/(2*√(5.88 × 10^(-7) * 7200)) ≈ 0.251 Now, we need to find the error function value erf(0.251). We can find this value using a numerical calculator or a table of values for the error function. After calculating, we get: erf(0.251) ≈ 0.2234
04

4. Calculate the temperature using the heat diffusion equation

Now, we plug in the values into the heat diffusion equation: T(x,t) = 288 + (268-288)*(1 - 0.2234) T(x,t) ≈ 288 - 20 * 0.7766 T(x,t) ≈ 277.47 K
05

5. Convert the final temperature back to Celsius and choose the answer

Now, we convert the final temperature back to Celsius: T(x,t) = 277.47 K - 273.15 = 4.32 °C Based on the options given, the closest value to 4.32 °C is 3 °C. Therefore, the answer is: (d) 3 °C

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Most popular questions from this chapter

Aluminum wires \(4 \mathrm{~mm}\) in diameter are produced by extrusion. The wires leave the extruder at an average temperature of \(350^{\circ} \mathrm{C}\) and at a linear rate of \(10 \mathrm{~m} / \mathrm{min}\). Before leaving the extrusion room, the wires are cooled to an average temperature of $50^{\circ} \mathrm{C}\( by transferring heat to the surrounding air at \)25^{\circ} \mathrm{C}\( with a heat transfer coefficient of \)50 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$. Calculate the necessary length of the wire cooling section in the extrusion room.

A large cast iron container \((k=52 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\alpha=1.70 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\) ) with \(4-\mathrm{cm}\)-thick walls is initially at a uniform temperature of \(0^{\circ} \mathrm{C}\) and is filled with ice at \(0^{\circ} \mathrm{C}\). Now the outer surfaces of the container are exposed to hot water at $55^{\circ} \mathrm{C}$ with a very large heat transfer coefficient. Determine how long it will be before the ice inside the container starts melting. Also, taking the heat transfer coefficient on the inner surface of the container to be $250 \mathrm{~W} / \mathrm{m}^{2}\(, \)\mathrm{K}$, determine the rate of heat transfer to the ice through a \(1.2-\mathrm{m}\)-wide and \(2-\mathrm{m}\)-high section of the wall when steady operating conditions are reached. Assume the ice starts melting when its inner surface temperature rises to $0.1^{\circ} \mathrm{C}$.

A hot dog can be considered to be a \(12-\mathrm{cm}\)-long cylinder whose diameter is \(2 \mathrm{~cm}\) and whose properties are $\rho=980 \mathrm{~kg} / \mathrm{m}^{3}\(, \)c_{p}=3.9 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}, k=0.76 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\(, and \)\alpha=2 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}\(. A hot dog initially at \)5^{\circ} \mathrm{C}\( is dropped into boiling water at \)100^{\circ} \mathrm{C}$. The heat transfer coefficient at the surface of the hot dog is estimated to be \(600 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the hot \(\operatorname{dog}\) is considered cooked when its center temperature reaches \(80^{\circ} \mathrm{C}\), determine how long it will take to cook it in the boiling water. Solve this problem using the analytical one-term approximation method.

What are the factors that affect the quality of frozen fish?

A thick wood slab \((k=0.17 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\alpha=1.28 \times\) \(10^{-7} \mathrm{~m}^{2} / \mathrm{s}\) ) that is initially at a uniform temperature of \(25^{\circ} \mathrm{C}\) is exposed to hot gases at \(550^{\circ} \mathrm{C}\) for a period of \(5 \mathrm{~min}\). The heat transfer coefficient between the gases and the wood slab is $35 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$. If the ignition temperature of the wood is \(450^{\circ} \mathrm{C}\), determine if the wood will ignite.

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