To illustrate the effect of ice on the aluminum cooling plate, consider the data shown in the drawing below.
The length of the ice block is L1 = 0.00476 m and its temperature is T1 = -11.9 °C. The aluminum block is L2 = 0.00131 m long and has a temperature of T2 = -23.8 °C. Ignore any limitations due to significant figures. Calculate the heat per second per square meter that is conducted through the ice-aluminum combination. Do not enter unit.
I'm not able to submit the URL for the picture so here's my best attempt at drawing it
|--------|--|
T1 | ice | | <---aluminum
|_________|___| T2
---L1---|-L2-
the picture got messed up. theres a block of ice on the left and the strip of aluminum on the right. the base of the ice is L1 and the base of the aluminum is L2. T1 is the ice and T2 is the aluminum. Please help me with this problem!!!!
Isn't this a simple heat flow problem.
Boundry coditions: At the ice/Al interface, temp 0deg. At the right side of the aluminum, Temp is T2. Heat flow through the ice, and then in aluminum will be equal. You cannot assume anything about the temperature of the ice/aluminum interface.
You need the thermal conductivity of ice, and aluminum, look them up.
Then start with the heat flow through the ice (deltaTemp T1-Tinterface), setting it equal to heat flow through the aluminum, (deltatemp Tinerface-T2). Solve for the unknown Temp at interface, then you can solve for heat flow.
To calculate the heat per second per square meter that is conducted through the ice-aluminum combination, we can use the formula for heat conduction:
Q = k * A * (T1 - T2) / d
Where:
Q is the heat conducted per second
k is the thermal conductivity of the material
A is the cross-sectional area
T1 and T2 are the temperatures on either side of the material
d is the thickness of the material
In this case, the ice and aluminum are in direct contact, so the thermal conductivity of the combination can be calculated using the formula for thermal resistances in series:
1/keff = 1/kice + L1/(kice*A) + L2/(kaluminum*A)
Where:
keff is the effective thermal conductivity of the combination
kice is the thermal conductivity of ice
kaluminum is the thermal conductivity of aluminum
L1 and L2 are the lengths of the ice and aluminum blocks, respectively
Now, we can calculate the heat per second per square meter (Q/A) using the known values:
1. Calculate the effective thermal conductivity of the combination (keff):
1/keff = 1/kice + L1/(kice*A) + L2/(kaluminum*A)
1/keff = 1/thermal conductivity of ice + L1/(thermal conductivity of ice * A) + L2/(thermal conductivity of aluminum * A)
keff = 1 / (1/thermal conductivity of ice + L1/(thermal conductivity of ice * A) + L2/(thermal conductivity of aluminum * A))
2. Calculate the heat per second per square meter (Q/A):
Q/A = keff * (T1 - T2) / (L1 + L2)
Note: Make sure to use the appropriate thermal conductivity values for ice and aluminum and convert the lengths (L1 and L2) into meters.
I hope this helps! Let me know if you have any further questions.
To calculate the heat per second per square meter conducted through the ice-aluminum combination, we need to use the formula for the rate of heat conduction. The formula is given by:
Q/t = k * A * ΔT / L
where:
Q/t is the heat conducted per second,
k is the thermal conductivity of the material (in this case, ice and aluminum),
A is the area of contact between the two materials,
ΔT is the temperature difference between the two materials, and
L is the thickness of the combination.
First, let's find the area of contact A between the ice and aluminum. Since the ice block is longer, we will use its length:
A = L1 * 1 (assuming the width of the ice block is 1)
Next, let's find the temperature difference ΔT between the two materials:
ΔT = T2 - T1
Now, we need to find the thermal conductivity of ice (k_ice) and aluminum (k_aluminum). The thermal conductivity values are typically given in units of Watts per meter-Kelvin (W/m-K). For this calculation, we will assume the thermal conductivities as follows:
k_ice = 2.2 W/m-K
k_aluminum = 205 W/m-K
Plug in the values into the formula:
Q/t = (k_ice * A * ΔT) / L
Finally, substitute the given values into the equation:
Q/t = (2.2 * (0.00476 * 1) * (T2 - T1)) / (0.00131 + 0.00476)
Evaluating this expression will give you the heat per second per square meter that is conducted through the ice-aluminum combination.