91Ó°ÊÓ

A. \(83 \mathrm{~mm}-\mathrm{high}\) Styrofoam cup has \(1.5 \mathrm{~mm}\)-thick walls and is filled with \(180 \mathrm{ml}\) coffee at \(80^{\circ} \mathrm{C}\) and covered with a lid. The outside diameter of the cup varies from \(45 \mathrm{~mm}\) at its base to \(73 \mathrm{~mm}\) at its top. The ambient air is at \(24^{\circ} \mathrm{C}\), and the combined convective and radiative heat transfer coefficient for the outside of the cup is estimated to be \(10 \mathrm{~W} / \mathrm{m} \mathrm{K}\). (i) Determine the initial rate of heat loss through the side walls of the cup and the corresponding temperature of the outer surface. (ii) Estimate the time for the coffee to cool to \(60^{\circ} \mathrm{C}\) if the average of the heat fluxes through the lid and base are taken to be equal to the flux through the side walls. Comment on the significance of your answer to actual cooling rates experienced at the morning coffee break. Take \(k=0.035 \mathrm{~W} / \mathrm{m} \mathrm{K}\) for the Styrofoam, and \(\rho=985\) \(\mathrm{kg} / \mathrm{m}^{3}, c_{p}=4180 \mathrm{~J} / \mathrm{kg} \mathrm{K}\) for the coffee.

Step by step solution

01

Calculate Surface Area of the Cup's Side Walls

The cup is a frustum of a cone, so we will use the formula for the lateral surface area of a frustum: \[ A = \pi (r_1 + r_2)\sqrt{(r_2 - r_1)^2 + h^2} \] where \( r_1 = 22.5 \mathrm{~mm} \), \( r_2 = 36.5 \mathrm{~mm} \), and \( h = 83 \mathrm{~mm} \). First, convert all dimensions to meters: \( r_1 = 0.0225 \mathrm{~m} \), \( r_2 = 0.0365 \mathrm{~m} \), and \( h = 0.083 \mathrm{~m} \). Calculate \( A \).

02

Determine Thermal Resistance of the Styrofoam Wall

The thermal resistance for conduction through a cylindrical annulus is given by \[ R = \frac{\ln\left(\frac{r_2}{r_1}\right)}{2 \pi L k} \] where \( L = 0.083 \mathrm{~m} \) and \( k = 0.035 \mathrm{~W/mK} \). Calculate \( R \) using the given dimensions.

03

Calculate Heat Loss Rate through Side Walls

Using the formula for heat transfer rate \( Q = \Delta T / R \), where \( \Delta T = 80^{\circ}C - 24^{\circ}C = 56^{\circ}C \), calculate the initial rate of heat loss \( \dot{Q} \). Substitute the previously calculated thermal resistance.

04

Determine Outer Surface Temperature of the Cup

Assume uniform heat flow and solve for the outer surface temperature \( T_s \) using the equation \( Q = hA(T_s - 24) \), where \( h = 10 \mathrm{~W/m}^2\mathrm{K} \), and \( A \) is the side wall area. Ensure consistency with \( \dot{Q} \) calculated in Step 3.

05

Compute Time for Coffee Cooling

Using the formula \( Q = mc\Delta T \), where \( m = 0.180 \mathrm{~kg} \), \( c = 4180 \mathrm{~J/kgK} \), and \( \Delta T = 80 - 60 = 20^{\circ}C \), solve for the energy \( Q \) needed to cool the coffee. Assuming constant heat flux calculated above, determine time \( t \) using \( t = Q/\dot{Q} \).

06

Comment on Cooling Rate

Discuss how assumptions about uniform heat loss through the side walls, base, and lid affect the real cooling rate of coffee during a break. Consider factors such as actual heat transfer coefficients and environmental conditions.

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Key Concepts

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

Thermal Resistance

When thinking about why your warm coffee cools so quickly, one key aspect to consider is thermal resistance. This is much like how electrical resistance works, but for heat. Thermal resistance is about how well a material impedes the flow of heat. In the case of the Styrofoam cup, its walls provide some resistance to the heat escaping.

For conduction through a cylindrical annulus, like the walls of the cup, the thermal resistance can be calculated as:\[R = \frac{\ln\left(\frac{r_2}{r_1}\right)}{2 \pi L k}\]

• \(r_1\) and \(r_2\) are the inner and outer radii of the cup walls.
• \(L\) is the height of the cup.
• \(k\) is the thermal conductivity, a measure of how well the material conducts heat.

The greater the thermal resistance, the slower the heat loss. It's why insulating materials like Styrofoam are so effective in keeping your drinks hot.

Heat Loss Rate

Next, we delve into how swiftly heat leaves your coffee, known as the heat loss rate. In simple terms, this is how fast the heat moves from the hot coffee through the cup into the cooler air outside.

The formula used here is:\[\dot{Q} = \frac{\Delta T}{R}\]

• \(\dot{Q}\) represents the rate of heat loss through the cup's sidewalls.
• \(\Delta T\) is the temperature difference between the coffee and the surrounding air.
• \(R\) is the previously calculated thermal resistance.

Think of this concept like opening a window on a cold day, the bigger the window (or lower the resistance), the faster the heat escapes. For the cup, ensuring that its thermal resistance is high means that the heat loss rate is slower, keeping your coffee warm for longer.

Cooling Time Calculation

Now, let’s look at how long it takes for your steaming coffee to cool to a more sippable temperature of 60°C from an initial temperature of 80°C. To find this, we use a basic energy balance equation, considering the coffee's mass and specific heat capacity.

The necessary energy to cool the coffee is given by:\[Q = mc\Delta T\]

• \(m\) is the mass of the coffee.
• \(c\) is the specific heat capacity.
• \(\Delta T\) is the change in temperature (20°C in this case).

Subsequently, we calculate the time it takes using the constant heat loss rate:\[ t = \frac{Q}{\dot{Q}}\]This represents the time required for your coffee to cool under the constant heat loss rate. However, real-world conditions like stirring or a breeze can significantly affect this calculation.

Thermal Conductivity

Lastly, let’s understand thermal conductivity, another crucial player in the heat game. This property indicates how easily a material can transfer heat. For the Styrofoam cup, the lower the thermal conductivity, the better it is at insulating.

In the context of the cup exercise, the Styrofoam's thermal conductivity is denoted by \(k = 0.035 \, \text{W/mK}\). A low \(k\) value suggests resistance to heat flow, thereby keeping the heat from escaping quickly. This is why materials with low conductivity are often used as insulation.

• Materials with high thermal conductivity make poor insulators but are great for heat exchangers.
• Conversely, low thermal conductivity equals better insulation properties.

So, the next time you wrap your hands around a Styrofoam cup, know that its low thermal conductivity is what’s keeping your drink warmer for longer.