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When we talk about thermal conductivity, we're discussing how well a material can conduct heat. Imagine it as a highway for heat flow. Some materials let heat zoom through quickly, while others slow it down. The thermal conductivity (\(k\)) measures this capability. In our exercise, the heat-shield tile from the space shuttle has a thermal conductivity of 0.065 \(\text{J/(s}\cdot\text{m}\cdot\text{C}^{\circ})\), meaning it's not an excellent conductor compared to metals like copper but does allow some heat to pass.If you think about it: - Higher thermal conductivity means better heat conduction. - Lower thermal conductivity suggests insulation properties.The thermal conductivity is used in the formula for heat transfer: \[Q = \frac{k \cdot A \cdot (T_2 - T_1) \cdot t}{d}\]Here, we multiply the thermal conductivity by the area, the temperature difference, and the time, then divide by the thickness. Knowing this, you can see how materials with high thermal conductivity transfer heat effectively over time.

Temperature change, symbolized as \(\Delta T\), tells us how much the temperature of a substance increases or decreases. It's essential in understanding how substances react when they gain or lose heat. In the shuttle tile example, the importance is in the difference between the outer and inner surfaces. That difference (\(T_2 - T_1\)) is 1130\(^{\circ}C\), a significant amount that drives heat across the tile.This concept is crucial when analyzing:

• how much heat energy is required to cause a change in temperature,
• the efficiency of heat transfer,
• and the impact on substances receiving heat.

By calculating the temperature difference, we ensure that the tile can handle extreme conditions, simulating how it protects the spacecraft by allowing only some heat to pass through.

Specific heat capacity (\(c\)) tells us how much heat is needed to raise the temperature of a given mass by one-degree Celsius. Each material has its own specific heat capacity, indicating its ability to absorb heat.For water, a commonly used reference example, the specific heat capacity is \(4.186 \text{J/g}\cdot\text{C}^{\circ}\). This means that to raise 1 gram of water by 1\(^{\circ}C\), you need 4.186 Joules of energy.In the context of the exercise:

• We've determined that 220.95\(J\) of heat was available from the tile,
• Water's specific heat capacity shows us how this heat changes its temperature.

By using the formula \(Q = m \cdot c \cdot \Delta T\), we rearrange it to solve for \(\Delta T\) and find out how much the water's temperature increases after receiving that specific heat. Here, it resulted in a minor change of about 0.0263\(^{\circ}C\), showing water's resilient heat holding capacity.