A polystyrene ice chest contains 3 kg of ice at 0 degrees Celsius. Triangle H for melting = 3.34 x 10 to the 5 power J/kg p = 1000 kg/m cubed, CP =4190 J/ (kg K). Heat transfer is limited by conduction (k=0.06W/m K)) from the outside wall at 30 degrees Celsius. This means that temperature within the ice chest is not a function of position. Assume the heat flux (J/(sec m squared) is the same through all walls: top, sides, bottom. The walls have thickness = 4 cm and total surface area = 2400 cm squared.
The water temperature increases after all the ice has melted. Assume that at any time the temperature profile across the wall is linear.
Calculate dTwater/dt (degrees Celsius/sec) when T water = 15 degrees Celsius.
To calculate dTwater/dt (the rate of change of water temperature with respect to time) when Twater = 15 degrees Celsius, we need to determine the heat transfer rate happening within the ice chest.
First, let's calculate the total heat transfer rate through the walls of the ice chest using the formula:
Qdot = k * A * dT / L
Where:
Qdot is the heat transfer rate (in watts)
k is the thermal conductivity of the wall material (in W/mK)
A is the total surface area through which heat is transferred (in m^2)
dT is the temperature difference between the inside and outside of the wall (in degrees Celsius)
L is the thickness of the wall (in meters)
Given:
k = 0.06 W/mK
A = 2400 cm^2 = 0.24 m^2
dT = (30 - 0) = 30 degrees Celsius (since the temperature inside is not a function of position)
L = 4 cm = 0.04 m
Substituting these values into the formula, we can calculate the heat transfer rate:
Qdot = 0.06 * 0.24 * 30 / 0.04
= 0.06 * 0.24 * 30 / 0.04
= 1.08 W
Next, we need to calculate the mass of the ice using the equation:
m = density * volume
Given:
density of ice = 1000 kg/m^3
volume of ice = 3 kg (since there are 3 kg of ice)
m = 1000 * 3
= 3000 kg
Now, let's determine the heat required to melt the ice:
Q = m * Hf
Where:
Q is the heat required to melt the ice (in joules)
m is the mass of the ice (in kilograms)
Hf is the heat of fusion (in joules/kg)
Given:
Hf = 3.34 x 10^5 J/kg
Q = 3000 * (3.34 x 10^5)
= 1.002 x 10^9 J
Since the heat is transferred from the walls to melt the ice, the rate of change of temperature of the water is given by the equation:
dTwater/dt = Qdot / (m * Cp)
Where:
dTwater/dt is the rate of change of water temperature with respect to time (in degrees Celsius/second)
Qdot is the heat transfer rate (in watts)
m is the mass of the water being heated (in kilograms)
Cp is the specific heat capacity of water (in joules/(kilogram * degree Celsius))
Given:
Twater = 15 degrees Celsius
Cp = 4190 J / (kg * K)
First, we need to find the mass of the water by converting the ice mass into water mass using the equation:
m = mi * (1 - Hf / Cp * (T - Ti))
Where:
m is the mass of the water (in kilograms)
mi is the initial mass of the ice (in kilograms)
Hf is the heat of fusion (in joules/kg)
Cp is the specific heat capacity of water (in joules/(kilogram * degree Celsius))
T is the final temperature of the water (in degrees Celsius)
Ti is the initial temperature of the ice (in degrees Celsius)
Given:
mi = 3 kg
Ti = 0 degrees Celsius (since the ice starts at 0 degrees Celsius)
m = 3 * (1 - (3.34 x 10^5) / (4190 * (15 - 0)))
= 2.445 kg
Finally, we can calculate dTwater/dt:
dTwater/dt = Qdot / (m * Cp)
= 1.08 / (2.445 * 4190)
= 0.000063 degrees Celsius/second
Therefore, the rate of change of water temperature when Twater = 15 degrees Celsius is approximately 0.000063 degrees Celsius/second.