/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 46 A Styrofoam box has a surface ar... [FREE SOLUTION] | 91影视

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Chapter 11: Problem 46

A Styrofoam box has a surface area of \(0.80 \mathrm{~m}^{2}\) and a wall thickness of \(2.0 \mathrm{~cm}\). The temperature of the inner surface is \(5.0^{\circ} \mathrm{C}\), and the outside temperature is \(25^{\circ} \mathrm{C}\). If it takes \(8.0 \mathrm{~h}\) for \(5.0 \mathrm{~kg}\) of ice to melt in the container, determine the thermal conductivity of the Styrofoam.

Short Answer

Expert verified
The thermal conductivity of the Styrofoam is \(0.0725 \, W/m \cdot K\).

Step by step solution

01

Compute the heat (Q) needed to melt the ice

We are given that 5.0 kg of ice melts in 8.0 hr. From the latent heat formula, we can compute the heat required to melt this ice. The formula we'll use is \( Q = m \cdot L_f \), where \(m = 5.0 \, kg\) is the mass of ice and \(L_f = 334 \, kJ/kg\) is the latent heat of fusion for ice. Thus, \(Q = 5.0 \, kg \cdot 334 \, kJ/kg = 1670 \, kJ\), which we convert to Joules because the formula for heat transfer requires SI units. That gives us \(Q = 1670 \, kJ \cdot 1000 \, J/kJ = 1.67 \times 10^6 \, J\).
02

Calculate the rate of heat transfer

The heat Q has been transferred over a time period of 8.0 hr. We calculate the rate of heat transfer (P) using the formula \(P = Q / t\), where \(t = 8.0 \, hr = 8.0 \, hr \cdot 3600 \, s/hr = 28800 \, s\). Thus, \(P = 1.67 \times 10^6 \, J / 28800 \, s = 58.0 \, J/s\).
03

Find the temperature difference

The temperature difference \(\Delta T\) across the Styrofoam is the exterior temperature minus the interior temperature. That gives us \( \Delta T = T_{outside} - T_{inside} = 25.0 ^{\circ}C - 5.0^{\circ}C = 20.0^{\circ}C\). However, since we need \(\Delta T\) in Kelvin for our next step, and since a change of 1 degree Celsius is equal to a change of 1 Kelvin, \(\Delta T = 20.0 K\).
04

Use the formula for heat transfer to solve for the thermal conductivity \(k\)

The formula for heat transfer \(P\) through a wall is \(P = kA\Delta T/d\), where \(A = 0.80 \, m^2\) is the surface area of the Styrofoam, \(\Delta T = 20.0 K\) is the temperature difference across the Styrofoam, and \(d = 2.0 \, cm = 2.0 \, cm/100 = 0.020 \, m\) is the thickness of the Styrofoam. We rearrange the formula to solve for the thermal conductivity: \(k = Pd/A\Delta T = 58.0 \, J/s * 0.02 \, m / (0.80 \, m^2 * 20.0 K) = 0.0725 \, W/m \cdot K \).

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Most popular questions from this chapter

\(\mid \mathbf{C}\) A \(3.00-g\) lead bullet at \(30.0^{\circ} \mathrm{C}\) is fired at a speed of \(2.40 \times 10^{2} \mathrm{~m} / \mathrm{s}\) into a large, fixed block of ice at \(0{ }^{\circ} \mathrm{C}\), in which it becomes embedded. (a) Describe the energy transformations that occur as the bullet is cooled. What is the final temperature of the bullet? (b) What quantity of ice melts?

Liquid nitrogen has a boiling point of \(77 \mathrm{~K}\) and a latent heat of vaporization of \(2.01 \times 10^{5} \mathrm{~J} / \mathrm{kg} . \mathrm{A} 25-\mathrm{W}\) electric heating element is immersed in an insulated vessel containing \(25 \mathrm{~L}\) of liquid nitrogen at its boiling point. (a) Describe the energy transformations that occur as power is supplied to the heating element. (b) How many kilograms of nitrogen are boiled away in a period of \(4.0\) hours?

The excess internal energy of metabolism is exhausted through a variety of channels, such as through radiation and evaporation of perspiration. Consider another pathway for energy loss: moisture in exhaled breath. Suppose you breathe out \(22.0\) breaths per minute, each with a volume of \(0.600 \mathrm{~L}\). Suppose also that you inhale dry air and exhale air at \(37^{\circ} \mathrm{C}\) containing water vapor with a vapor pressure of \(3.20 \mathrm{kPa}\). The vapor comes from the evaporation of liquid water in your body. Model the water vapor as an ideal gas. Assume its latent heat of evaporation at \(37^{\circ} \mathrm{C}\) is the same as its heat of vaporization at \(100^{\circ} \mathrm{C}\). Calculate the rate at which you lose energy by exhaling humid air.

Three liquids are at temperatures of \(10^{\circ} \mathrm{C}, 20^{\circ} \mathrm{C}\), and \(30^{\circ} \mathrm{C}\), respectively. Equal masses of the first two liquids are mixed, and the equilibrium temperature is \(17^{\circ} \mathrm{C}\). Equal masses of the second and third are then mixed, and the equilibrium temperature is \(28^{\circ} \mathrm{C}\). Find the equilibrium temperature when equal masses of the first and third are mixed.

In the summer of 1958 in St. Petersburg, Florida, a new sidewalk was poured near the childhood home of one of the authors. No expansion joints were supplied, and by mid-July the sidewalk had been completely destroyed by thermal expansion and had to be replaced, this time with the important addition of expansion joints! This event is modeled here. A slab of concrete \(4.00 \mathrm{~cm}\) thick, \(1.00 \mathrm{~m}\) long, and \(1.00 \mathrm{~m}\) wide is poured for a sidewalk at an ambient temperature of \(25.0^{\circ} \mathrm{C}\) and allowed to set. The slab is exposed to direct sunlight and placed in a series of such slabs without proper expansion joints, so linear expansion is prevented. (a) Using the linear expansion equation (Eq. 10.4), eliminate \(\Delta L\) from the equation for compressive stress and strain (Eq. 9.3). (b) Use the expression found in part (a) to eliminate \(\Delta T\) from Equation \(11.3\), obtaining a symbolic equation for thermal energy transfer \(Q\). (c) Compute the mass of the concrete slab given that its density is \(2.40 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}\). (d) Concrete has an ultimate compressive strength of \(2.00 \times 10^{7} \mathrm{~Pa}\), specific heat of \(880 \mathrm{~J} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\), and Young's modulus of \(2.1 \times 10^{10} \mathrm{~Pa}\). How much thermal energy must be transferred to the slab to reach this compressive stress? (e) What temperature change is required? (f) If the Sun delivers \(1.00 \times 10^{3} \mathrm{~W}\) of power to the top surface of the slab and if half the energy, on the average, is absorbed and retained, how long does it take the slab to reach the point at which it is in danger of cracking due to compressive stress?
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