To solve for the heat loss due to conduction, we’ll use the formula for heat conduction through a material:

$Q = \frac{k \cdot A \cdot \Delta T}{d}$

where:

• $Q$ is the heat transfer per unit time (heat loss rate, in cal/s),
• $k$ is the thermal conductivity of the material ($0.0002 \, \text{cal/s} \cdot \text{cm} \cdot ^\circ\text{C}$),
• $A$ is the surface area of the cube (in $\text{cm}^2$),
• $\Delta T$ is the temperature difference between the coffee and the surroundings ($85^\circ\text{C} - 28^\circ\text{C} = 57^\circ\text{C}$),
• $d$ is the thickness of the insulating material ($0.5 \, \text{cm}$).

Step 1: Calculate the Surface Area

Since the cup is shaped like a cube with a side length of $9.3 \, \text{cm}$, its surface area $A$ is:

$A = 6 \cdot \text{side}^2 = 6 \cdot (9.3 \, \text{cm})^2 = 6 \cdot 86.49 = 518.94 \, \text{cm}^2$

Step 2: Calculate the Heat Loss Rate in Calories per Second

Substitute the known values into the formula:

$Q = \frac{0.0002 \cdot 518.94 \cdot 57}{0.5}$

Calculating the numerator:

$0.0002 \cdot 518.94 \cdot 57 = 5.938476$

Now divide by $0.5$:

$Q = \frac{5.938476}{0.5} = 11.876952 \, \text{cal/s}$

Step 3: Convert Calories per Second to Joules per Second

Since $1 \, \text{cal} = 4.184 \, \text{J}$:

$Q = 11.876952 \times 4.184 = 49.692 \, \text{J/s}$

Final Answer

The heat loss due to conduction is approximately:

$Q \approx 49.69 \, \text{J/s}$

Would you like a breakdown of each step in further detail, or do you have any questions?

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Here are some related questions to explore further:

1. How would the heat loss change if the cup's thickness increased?
2. What effect would a lower surrounding temperature have on the heat loss rate?
3. How does changing the thermal conductivity affect the rate of heat loss?
4. If the shape were cylindrical, how would we approach calculating surface area?
5. How does heat loss due to conduction compare to convection in an insulated container?

Tip: When dealing with conduction problems, always check units to ensure consistency, especially during conversions from calories to joules.