/*! 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 66 Two large blocks of different ma... [FREE SOLUTION] | 91Ó°ÊÓ

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Two large blocks of different materials, such as copper and concrete, have been sitting in a room \(\left(23^{\circ} \mathrm{C}\right)\) for a very lone time. Which of the two blocks, if either, will feel colder to the touch? Assume the blocks to be seminfinite solids and your hand to be at a temperaturt of \(37^{\circ} \mathrm{C}\).

Short Answer

Expert verified
Copper feels colder due to higher thermal conductivity and heat absorption capacity.

Step by step solution

01

Understand the Problem

We need to determine which block feels colder when touched, knowing that both blocks have been sitting in a room at thermal equilibrium (23°C). The materials have different thermal properties, and our hand is at 37°C.
02

Define Relevant Concepts

When we touch an object, the sensation of 'coldness' depends on the rate of heat transfer from our hand to the object. This rate is influenced by the material's thermal conductivity, specific heat, and density.
03

Identify Thermal Conductivity

Copper has a high thermal conductivity, meaning it transfers heat quickly. In contrast, concrete has a lower thermal conductivity, making the heat transfer slower.
04

Consider Heat Absorption Capacity

Copper's higher density and specific heat compared to concrete mean it can absorb more heat from your hand. This implies that copper will draw heat away more effectively, enhancing the sensation of coldness.
05

Assess the Thermal Touch Sensation

Since copper conducts and absorbs heat more rapidly, it will feel colder than concrete when touched. The heat is quickly drawn away from the skin, creating a greater sensation of cold.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Heat Transfer
Heat transfer is the process by which heat energy moves from one object to another. This movement can occur in three main ways: conduction, convection, and radiation. When we touch something, heat transfer by conduction is the key factor because heat energy flows directly from one material to another. In the context of our original exercise, when our hand touches any object, like the copper or concrete blocks, heat is transferred from our warmer skin to the cooler blocks. This occurs because heat naturally flows from high-temperature areas to low-temperature areas until equilibrium is reached. The rate of heat transfer affects how we perceive the temperature of an object. Materials with high thermal conductivity, like copper, transfer heat more quickly from our hand, creating a sensation of rapid cooling. In contrast, materials with low thermal conductivity, such as concrete, transfer heat slowly, thereby feeling less cold initially.
Specific Heat
Specific heat is a property of a material that indicates how much heat energy it takes to raise the temperature of a given mass by one degree Celsius. It tells us about a material's ability to absorb and store heat. Materials with high specific heat capacity require more energy to change their temperature. This means they can absorb more heat without rising in temperature as quickly. In our example, copper has a high specific heat, so it can absorb more heat from our hand compared to concrete. This higher absorption rate means that copper will remove heat from our hand more efficiently, enhancing the sensation of cold when we touch it.
Thermal Equilibrium
Thermal equilibrium is a state where there is no net flow of heat energy between objects in contact; they exist at the same temperature. When objects reach thermal equilibrium, they do not exchange heat energy, creating a stable condition. In the exercise, both the copper and concrete blocks are initially at thermal equilibrium with the room, measured at 23°C. Our hand, however, is warmer at 37°C. When we touch either block, the heat from our hand will transfer to the block until both are at a temperature that reflects a new state of thermal equilibrium. Copper, being a good conductor of heat, quickly draws heat away from our hand, leading to a more pronounced feeling of cold before equilibrium is achieved. Concrete, with its lower thermal conductivity, reaches equilibrium with our hand at a slower pace, making it feel warmer in comparison.

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Most popular questions from this chapter

The convection coefficient for flow over a solid sphere may be determined by submerging the sphere, which is initially at \(25^{\circ} \mathrm{C}\), into the flow, which is at \(75^{\circ} \mathrm{C}\), and measuring its surface temperature at some time during the transient heating process. (a) If the sphere has a diameter of \(0.1 \mathrm{~m}\), a thermal conductivity of \(15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and a thermul diffusivity of \(10^{-5} \mathrm{~m}^{2} / \mathrm{s}\), at what time will a surface tempertiture of \(60^{\circ} \mathrm{C}\) be recorded if the convection coefficient is \(300 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) ? (b) Assess the effect of thermal diffusivity on the thermal response of the material by computing center and surface temperature histories for \(\alpha=10^{-6}\). \(10^{-5}\), and \(10^{-4} \mathrm{~m}^{2} / \mathrm{s}\). Plot your results for the perid \(0 \leq t \leq 300 \mathrm{~s}\). In a similar manner, assess tte effect of themal conductivity by considering vak ues of \(k=1.5,15\), and \(150 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).

A support rod \(\left(\mathrm{k}=15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \alpha=4.0 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\right)\) of diameter \(D=15 \mathrm{~mm}\) and length \(L=100 \mathrm{~mm}\) spans a channel whose walls are maintained at a temperature of \(T_{b}=300 \mathrm{~K}\). Suddenly, the rod is exposed to a cross flow of hot gaces for which \(T_{2}=600 \mathrm{~K}\) and \(h=75 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The channel walls are cooled and remain at \(300 \mathrm{~K}\). (a) Using an appropriate numerical technique, determine the thermal response of the rod to the convective heating. Plot the midspan remperature as a function of elapsed time. Using an appropriate analytical model of the rod, determine the steady. state temperature distribution and compare the result with that obtained numerically for very long clapsed times. (b) After the rod has reached steady- state condisiss the flow of hot gases is suddenly termirated art the rod cools by free convection to ambient ans \(T_{w}=300 \mathrm{~K}\) and by radiafion exchange with kare surroundings at \(T_{\text {wr }}=300 \mathrm{~K}\). The free convetion coefficient can be expressed as \(h\left(\mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\right)=\mathrm{C}\) \(\Delta T^{k}\), where \(C=4.4 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}^{128}\) and \(n=0.18\) The emissivity of the rod is \(0.5\). Determire th subsequent thermal response of the rod. Pat th midspan temperature as a function of cooling tine. and determine the time required for the rod to raxt a safe-fo-fouch temperature of \(315 \mathrm{~K}\).

As permanent space stations increase in size, there is an attendant incrense in the amount of electrical power they dissipate. To keep station compartment temperatures from exceeding prescribed limits, it is necesary to transier the dissiputed heat to space. A novel heat rejection scheme that has been proposed for this purpose is termed a Liquid Droplet Radiator (T.DR). The heat is firs transferred to a high vacuum oil, which is then injected into outer space as a stream of mall droplets. The stream is allowed to traverse a disance \(L\). over which it cools by radiating energy to outer space at ubsolute zero temperiture. The droplets are then collected and routed back to the space station. Consider conditions for which droplets of emissivity \(E=0.95\) and diameter \(D=0.5 \mathrm{~mm}\) are injected at a temperature of \(T_{1}=500 \mathrm{~K}\) and a velocity of \(V=0.1 \mathrm{~m} / \mathrm{s}\). Properties of the oil are \(\rho=885 \mathrm{~kg} / \mathrm{m}^{3}, c=1900 \mathrm{~J} / \mathrm{kg}\). \(\mathrm{K}\), and \(k=0.145 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Assuming each drop to radiate to deep space at \(T_{\text {ar }}=0 \mathrm{~K}\), determine the distance \(L\) required for the droplets to impact the collector at a final temperature of \(T_{f}=300 \mathrm{~K}\). What is the amount of thermal energy rejecied by each droplet?

In a thin-slab, continuous casting process, molten steel leaves a mold with a thin solid shell, and the molien material solidifies as the slab is quenched by water jets en route to a section of rollers. Once fully solidified, the slab continues to cool as it is brought to an acceptable handling temperature. It is this portion of the process that is of interest. Consider a 200 - mm- thick solid slab of steel \((\rho=\) \(\left.7800 \mathrm{~kg} / \mathrm{m}^{3}, c=700 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, k=30 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right)\), initially at a uniform temperature of \(T_{i}=1400^{\circ} \mathrm{C}\). The slab is cooled at its top and bottom surfaces by water jets \(\left(T_{w}\right.\) \(\left.=50^{\circ} \mathrm{C}\right)\). which maintain an approximstely uniform convection coefficient of \(h=5000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) at both surfaces. Using a finite-difference solution with a spoce increment of \(\Delta x=1 \mathrm{~mm}\), determine the time required to cool the surface of the slab to \(200^{\circ} \mathrm{C}\). What is the corresponding temperature at the midplane of the slab? If the slab moves at a speed of \(V=15 \mathrm{~mm} / \mathrm{s}\), what is the required length of the cooling section?

A cold air chamber is proposed for quenching steel ball bearings of dinmeter \(D=0.2 \mathrm{~m}\) and initial temperature \(T_{i}=400^{\circ} \mathrm{C}\). Air in the chamber is maintained at \(-15^{\circ} \mathrm{C}\) by a refrigeration system, and the steel balls pass through the chamber on a conveyor belt. Optimum bearing production requires that \(70 \%\) of the initial thermal energy content of the ball above \(-15^{\circ} \mathrm{C}\) be removed. Radiation effects muy be neglected, and the convection heut transfer coefficient within the chamber is \(1000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Estimute the residence time of the balls within the chamber, und recommend a drive velocity of the conveyor. The following properties may be used for the steel: \(k=50 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \alpha=2 \times 10^{-3} \mathrm{~m}^{2} / \mathrm{s}\), and \(c=450 \mathrm{~J} / \mathrm{kg}-\mathrm{K}\).

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