A cubical piece of heat-shield-tile from the space shuttle measures 0.18 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 1050°C, while the inner surface is maintained at a temperature of 22°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 3.6 liters (3.6 kg) of liquid water, by how many Celsius degrees would the temperature of the water rise?
(a) Q=q•t =K•(ΔT/Δx) •A•t =
= 0.065•(1050-22)/0.18•(0.18)² •4•60 = =2887 J.
(b) Q = c•m• ΔT,
ΔT = Q/c•m = 2887/4180•3.6 = 0.2º.
how to find the c?
To find the amount of heat that flows 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 = heat (in joules)
k = thermal conductivity (in J/(s·m·C°))
A = surface area (in square meters)
ΔT = change in temperature (in Celsius degrees)
t = time (in seconds)
d = thickness of the tile (in meters)
Given:
k = 0.065 J/(s·m·C°)
A = (0.18 m)^2 = 0.0324 m^2
ΔT = 1050°C - 22°C = 1028°C
t = 4.0 minutes = 4.0 * 60 = 240 seconds
d = unknown
(a) To find the thickness of the tile:
Rearranging the formula, we have:
Q = (k * A * ΔT * t) / d
d = (k * A * ΔT * t) / Q
Substituting the given values:
d = (0.065 J/(s·m·C°) * 0.0324 m^2 * 1028°C * 240 seconds) / Q
Now, we need to calculate the value of Q.
Note: Q can be calculated using the same formula by substituting the given values.
(b) Once we have the value of Q, we can use the specific heat capacity of water to find the change in temperature.
The specific heat capacity of water is 4.186 J/(g·C°). We know that 1 liter of water is equal to 1 kg (since the density of water is 1 g/mL).
Given:
Q = unknown (from part a)
Specific heat capacity of water = 4.186 J/(g·C°)
Mass of water = 3.6 kg (or 3.6 liters)
Using the formula:
Q = m * c * ΔT
Where:
Q = heat (in joules)
m = mass (in grams)
c = specific heat capacity (in J/(g·C°))
ΔT = change in temperature (in Celsius degrees)
We need to solve for ΔT.
Let's calculate the values step-by-step.
To solve this problem, we can use the formula for heat transfer: Q = k * A * ΔT / Δx, where Q is the heat transferred, k is the thermal conductivity, A is the surface area, ΔT is the temperature difference, and Δx is the thickness of the material.
(a) To find the heat transferred from the outer to the inner surface of the tile in 4.0 minutes, we need to calculate Q.
First, let's find the surface area of the tile. Since it is a cube, all sides have the same length. So, A = (0.18 m)^2 * 6 = 0.1944 m^2.
Next, let's calculate the temperature difference. ΔT = (1050°C - 22°C) = 1028°C. Note that we need to convert this temperature difference to Kelvin since the formula requires temperature in Kelvin. So, ΔT = (1028°C + 273.15 K) = 1301.15 K.
The thickness of the tile is not given in the question. So, we cannot calculate the exact value of Δx. Hence, we assume a value for Δx, and the final answer will depend on the chosen value.
Now, let's assume a value of Δx. For simplicity, let's assume Δx is 0.01 m.
Using the given thermal conductivity (k = 0.065 J/(s·m·C°)), surface area (A = 0.1944 m^2), temperature difference (ΔT = 1301.15 K), and assumed thickness (Δx = 0.01 m), we can calculate the heat transferred (Q) as follows:
Q = (0.065 J/(s·m·C°)) * (0.1944 m^2) * (1301.15 K) / (0.01 m)
Q ≈ 1670 J
Therefore, approximately 1670 Joules of heat flow from the outer to the inner surface of the tile in 4.0 minutes.
(b) To find the temperature rise of water when it receives the transferred heat, we can use the formula: Q = mcΔT, where Q is the heat transferred, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the temperature change.
First, let's convert the volume of water to mass. We know that 1 liter of water has a mass of 1 kg. So, 3.6 liters of water has a mass of 3.6 kg.
The specific heat capacity of water (c) is approximately 4186 J/(kg·K).
Using the known values of Q (1670 J) and m (3.6 kg), we can calculate the temperature change (ΔT) as follows:
Q = mcΔT
1670 J = (3.6 kg) * (4186 J/(kg·K)) * ΔT
Now, let's solve for ΔT:
ΔT = 1670 J / (3.6 kg * 4186 J/(kg·K))
ΔT ≈ 0.111 K
Therefore, the temperature of the water would rise by approximately 0.111 degrees Celsius when it receives the transferred heat.