Question

A cubical piece of heat-shield-tile from the space shuttle measures 0.14 m on a side and has a thermal conductivity of 0.065 J/(s−m⋅C∘). The outer surface of the tile is heated to a temperature of 1210∘C, while the inner surface is maintained at a temperature of 19∘C. (a) How much heat flows from the outer to the inner surface of the tile in 4.0 minutes? (b) If this amount of heat were transferred to 3.1 liters ( 3.1 kg ) of liquid water, by how many Celsius degrees would the temperature of the water rise? (a) Number Units (b) Number Units

Answer 1

The heat flow from the outer to the inner surface of the tile in 4.0 minutes is 34,680 J.

Step-by-step explanation:

To determine the heat flow, we can use the formula for heat transfer:
`Q=k . A .` Δ
`T . t`, where
`Q` is the heat transferred,
`k` is the thermal conductivity,
`A` is the surface area, Δ
`T` is the temperature difference, and
`t` is time. The surface area of the cube is
`A = 6` ×
`(0.14 m)^(2)` as there are 6 faces in a cube. The temperature difference is Δ T= 1210°C - 19°C. Substituting these values into the formula gives Q = 0.065 J (s/cdoptm/cdopt °C)} × 6 ×
`(0.14m)^(2)` × (1210°C - 19°C) × (4.0 minutes × 60 s/min which equals 34,680 J.

Final answer:

If this heat were transferred to 3.1 liters (3.1 kg) of liquid water, the temperature of the water would rise by approximately 31.6°C

Step-by-step explanation:

Using the specific heat capacity of water (\(c = 4.186 \, \text{J/(g·°C)}\)), the formula \(Q = mc\Delta T\) can be used to find the temperature change of the water. Rearranging the formula to solve for \(\Delta T\) gives \(\Delta T = \frac{Q}{mc}\). Substituting \(Q = 34,680 \, \text{J}\), \(m = 3.1 \, \text{kg} = 3100 \, \text{g}\), and \(c = 4.186 \, \text{J/(g·°C)}\) gives \(\Delta T = \frac{34,680 \, \text{J}}{3100 \, \text{g} \times 4.186 \, \text{J/(g·°C)}} \approx 31.6°C\).

This calculation demonstrates that the heat transferred from the tile to the water would result in a temperature rise of approximately 31.6°C, assuming no heat losses to the surroundings.

Answer 2

The heat flow from the outer to the inner surface of the tile in 4.0 minutes is approximately 301.92 J. The temperature of the water would rise by approximately 22.91°C when this amount of heat is transferred.

Step-by-step explanation:

To calculate the amount of heat flowing from the outer to the inner surface of the tile in 4.0 minutes, we can use the formula:

Q = (k * A * ΔT * t) / d

Where:

• Q is the heat flow
• k is the thermal conductivity
• A is the surface area
• ΔT is the temperature difference
• t is the time
• d is the thickness

Substituting the given values into the formula, we get:

Q = (0.065 J/(s-m⋅C°) * (0.14 m)^2 * (1210°C - 19°C) * (4.0 minutes * 60 seconds/minute)) / (0.14 m)

Simplifying the equation, we find that the heat flow from the outer to the inner surface of the tile is approximately 301.92 J in 4.0 minutes.

To calculate the temperature rise of the water when this amount of heat is transferred, we can use the formula:

ΔT = Q / (m * c)

Where:

• ΔT is the temperature rise
• Q is the heat transferred
• m is the mass of the water
• c is the specific heat capacity of water (4.18 J/(g⋅C°))

Substituting the given values into the formula, we get:

ΔT = (301.92 J) / ((3.1 kg) * (4.18 J/(g⋅C°)))

Simplifying the equation, we find that the temperature rise of the water is approximately 22.91°C when this amount of heat is transferred.