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If a heated object of mass \(m\) is placed in a liquid that is maintained at a temperature \(T_{\ell},\) the temperature, \(T,\) of the object as a function of time, \(t,\) can be estimated using the relation $$\frac{\mathrm{d} T}{\mathrm{~d} t}=\frac{h}{m c}\left(T-T_{\ell}\right)$$ where \(c\) is the specific heat of the object and \(h\) is the heat transfer coefficient. In a typical application, the units of the variables are as follows: \(T\left({ }^{\circ} \mathrm{C}\right), t(\mathrm{~s}), m(\mathrm{~kg}), c\) \(\left(\mathrm{kJ} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\right),\) and \(T_{\ell}\left({ }^{\circ} \mathrm{C}\right) .\) In what units should \(h\) be expressed?

Short Answer

Expert verified
The units for \(h\) should be \(\text{kJ/} (\text{s} \cdot {}^{\circ}\text{C})\)."

Step by step solution

01

Analyzing the equation

The given differential equation is \(\frac{\mathrm{d} T}{\mathrm{~d} t}=\frac{h}{m c}\left(T-T_{\ell}\right)\). The term \(\frac{\mathrm{d} T}{\mathrm{~d} t}\) represents the rate of change of temperature with respect to time, which in terms of units is \(\text{ }^{\circ} \text{C} / \text{s}\).
02

Examine the right side expression

The right side of the equation \(\frac{h}{m c}\left(T-T_{\ell}\right)\) has an expression \((T-T_{\ell})\) that is simply a temperature difference, measured in degrees Celsius \(({}^{\circ} \text{C})\).
03

Equating units on both sides

For the equation to hold true, the units of \(\frac{h}{m c} (T - T_{\ell})\) must match the units of \(\frac{\mathrm{d} T}{\mathrm{~d} t}\), which are degrees Celsius per second \(({}^{\circ} \text{C} / \text{s})\). Thus, \(\frac{h}{m c}\) must have units of \(1/\text{s}\).
04

Solve for the units of h

Given that \(m\) is in kilograms \((\text{kg})\) and \(c\) is in kilojoules per kilogram per degree Celsius \((\text{kJ/}\text{kg} \cdot {}^{\circ} \text{C})\), we set \(\frac{h}{\text{kg} \cdot \text{kJ/}\text{kg} \cdot {}^{\circ}\text{C}} = \frac{1}{{}\text{s}}\). Solving for \(h\), \(h\) must be in \(\text{kJ/} (\text{s} \cdot {}^{\circ}\text{C})\).

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Key Concepts

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

Differential Equation
A differential equation is a type of equation that involves the derivatives of a function. In this context, it is used to express how the temperature of an object changes over time when placed in a liquid. The equation given is \(\frac{\mathrm{d} T}{\mathrm{d} t}=\frac{h}{m c}\left(T-T_{\ell}\right)\). Here, \(T\) represents the temperature of the object at time \(t\), and \(T_\ell\) is the constant temperature of the surrounding liquid.
  • \( \frac{\mathrm{d} T}{\mathrm{d} t} \): This is the derivative of \(T\) with respect to \(t\), indicating the rate at which the temperature changes.
  • The term \(\frac{h}{m c}(T-T_{\ell})\) describes how much influence the surrounding temperature has on the object.
Differential equations are powerful tools for modeling real-world situations where changes occur continuously over time, such as temperature variation.
Temperature Change
Temperature change refers to the difference between the initial and final temperatures during a process. In our equation, the difference \(T - T_\ell\) is critical. It shows the driving force for heat transfer, influencing how fast the temperature of the object approaches the liquid's temperature.- Initially, if the object's temperature is much higher than \(T_\ell\), \(T - T_\ell\) is large, leading to a higher rate of temperature change.- As the object's temperature nears \(T_\ell\), the difference \(T - T_\ell\) decreases, slowing the rate of temperature change.Understanding this concept helps us predict how quickly an object will reach thermal equilibrium with its surroundings.
Heat Transfer Coefficient
The heat transfer coefficient, denoted as \(h\), plays a pivotal role in dictating the rate of heat exchange between the object and the liquid. It essentially determines how effectively heat is transferred through a surface.- The larger the value of \(h\), the more efficient the heat transfer process. This means the object's temperature will change more quickly.- In terms of units, \(h\) must balance the equation in such a way that the units on both sides are consistent. Therefore, \(h\) is expressed in \(\text{kJ/} (\text{s} \cdot {}^{\circ}\text{C})\)Understanding \(h\) helps in designing processes and materials that ensure effective heat regulation in a given system.
Specific Heat
Specific heat, represented as \(c\), describes how much heat energy is needed to change the temperature of a unit mass by one degree Celsius. It is a material property that indicates the amount of energy required to raise the temperature.
  • Materials with high specific heat can absorb more energy without a significant change in temperature.
  • Low specific heat means temperature changes occur more readily with energy input or removal.
In our differential equation, \(c\) influences the thermal response of the object when placed in the liquid. If \(c\) is large, the object's temperature will change more slowly, indicating a higher thermal inertia. Knowing the specific heat of a material can be crucial in processes involving heating and cooling, as it affects energy calculations and time to reach desired temperatures.

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Most popular questions from this chapter

The Froude number, Fr, at any cross section of an open channel is defined by the relation $$\mathrm{Fr}=\frac{\bar{V}}{\sqrt{g D_{\mathrm{h}}}}$$ where \(\bar{V}\) is the average velocity, \(g\) is the acceleration due to gravity, and \(D_{\mathrm{h}}\) is the hydraulic depth. The hydraulic depth is defined as \(A / T,\) where \(A\) is the flow area and \(T\) is the top width of the flow area. (a) Show that Fr is dimensionless. (b) Determine the value of Fr in a trapezoidal channel that has a bottom width of \(3 \mathrm{~m}\), side slopes \(2.5: 1(\mathrm{H}: \mathrm{V}),\) an average velocity of \(0.4 \mathrm{~m} / \mathrm{s},\) and a flow depth of \(1.5 \mathrm{~m} .\)

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A hydraulic jump is a phenomenon associated with high-velocity flow of a liquid in an open channel in which the flow depth of the liquid suddenly changes from a lower depth, \(h_{1}\), to a higher depth, \(h_{2}\). When the flow is occurring in a horizontal rectangular channel, the relationship between \(h_{1}\) and \(h_{2}\) depends only on the lower-depth velocity of flow, \(V_{1}\), and the gravity constant, \(g\). Use dimensional analysis to determine the functional relationship between \(h_{2}\) and the influencing variables, expressed in terms of dimensionless groups. Identify any named conventional dimensionless groups that occur in this relationship.

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Consider the case where a particle of diameter \(d\) and density \(\rho_{\mathrm{s}}\) rests on the bottom of an open channel where a liquid with density \(\rho\) and viscosity \(\mu\) flows over the particle. The particle begins to move when the fluid velocity is \(V_{\mathrm{c}},\) and in this analysis, gravity, \(g,\) is a relevant variable. (a) Determine a dimensionless group of the given variables that measures the ratio of the fluid force tending to move the particle to the weight of the particle. Take into account that the particle does not weigh the same when submerged in a liquid as it does when not submerged. (b) Using this result, perform a dimensional analysis on the relevant variables to determine a functional relationship between dimensionless groups that can be used to analyze measurements in a scale model to measure the critical fluid velocity, \(V_{\mathrm{c}},\) to move a particle of a given size, \(d\). In your dimensional analysis, incorporate the fact that \(\rho_{\mathrm{s}}\) and \(g\) can be combined into the single variable \(\left(\rho_{\mathrm{s}}-\rho\right) g .(\mathrm{c})\) If a scale model is to be built that is twice the size of the actual system, what dimensionless group(s) would you make the same in the model and the prototype? If model measurements using a 3-mm-diameter particle show a critical velocity of \(25 \mathrm{~cm} / \mathrm{s}\), what is the corresponding critical velocity and particle size in the prototype?

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