Question:

A solid cylinder of aluminum (thermal conductivity = 237 W/m·K, Specific heat capacity = 0.897 J/g·K) with a radius of 5 cm and a length of 20 cm is initially at a temperature of 25°C. The cylinder is immersed in a water bath at a constant temperature of 100°C.

a) Calculate the time it takes for the center of the cylinder to reach a temperature of 75°C.

b) Calculate the rate at which heat is conducted through the cylinder when its center reaches a temperature of 75°C.

c) Calculate the thermal energy absorbed by the water bath during the time calculated in part a), assuming no heat losses to the surroundings.

Answer:

Given Data:

• Thermal conductivity of aluminum (k) = 237 W/m·K
• Specific heat capacity of aluminum (C_al) = 0.897 J/g·K
• Radius of the cylinder (r) = 5 cm = 0.05 m
• Length of the cylinder (L) = 20 cm = 0.2 m
• Initial temperature of the cylinder (T_initial) = 25°C
• Temperature of the water bath (T_water) = 100°C
• Temperature when the center of the cylinder reaches (T_final) = 75°C

a) To calculate the time it takes for the center of the cylinder to reach a temperature of 75°C, we need to use the equation for one-dimensional heat conduction:

Where:

• q is the rate of heat conducted through the cylinder (in watts)
• k is the thermal conductivity of aluminum (in W/m·K)
• A is the cross-sectional area of the cylinder (in m²)
• T_f is the final temperature (in K)
• T_i is the initial temperature (in K)
• L is the length of the cylinder (in m)

We can first calculate the cross-sectional area of the cylinder:

Converting the given temperatures to Kelvin: Initial temperature (T_initial) = 25°C + 273.15 = 298.15 K Final temperature (T_final) = 75°C + 273.15 = 348.15 K

Substituting the given data into the equation, we have:

Now, we can solve for q, the rate of heat conducted through the cylinder:

b) Now, to calculate the rate at which heat is conducted through the cylinder when its center reaches a temperature of 75°C, we use the formula:

Where:

• dT/dx is the change in temperature per unit length
• q is the rate of heat conducted through the cylinder (in watts)
• k is the thermal conductivity of aluminum (in W/m·K)
• A is the cross-sectional area of the cylinder (in m²)

Solving for dT/dx, we can rewrite the equation as:

c) Finally, to calculate the thermal energy absorbed by the water bath during the time calculated in part a), we use the formula:

Where:

• ΔQ is the thermal energy absorbed (in joules)
• m is the mass of the water (in grams)
• c is the specific heat capacity of water (in J/g·K)
• ΔT is the change in temperature of the water (in K)

To find the mass of the water, we can use the formula:

Where:

• ρ is the density of water (in g/cm³)
• V is the volume of water (in cm³)

Now, we can substitute the given values to find the thermal energy absorbed by the water bath:

The final answer requires us to plug in the calculated time value from part a) into the equation, as the time period during which the energy is absorbed.

Note: The specific heat capacity of water (c) is approximately 4.18 J/g·K, and the density of water (ρ) is 1 g/cm³.