Sunlight of intensity 800 W m^-2 is captured by a tank containing 100 kg of water with an effici

step 1 First, we find the volume of the tank. So we take the entire volume and then subtract the hollow

Sunlight of intensity 800 W m^-2 is captured by a tank...

Sunlight of intensity $800 \mathrm{W} \mathrm{m}^{-2}$ is captured by a tank containing $100 \mathrm{kg}$ of water with an efficiency of $80 \%$. The tank is rectangular in shape and has dimensions $1.0 \times 1.0 \times 0.10 \mathrm{m}^{3}$. It has walls of thickness $5.0 \mathrm{mm} .$ The surrounding air has a temperature of $20^{\circ} \mathrm{C}$. Assume that the tank is well insulated from all sides except the top surface (of area $1.0 \mathrm{m}^{2}$ ). The material of the tank has a thermal conductivity of $k=0.30 \mathrm{Wm}^{-1} \mathrm{K}^{-1},$ its density is $1200 \mathrm{kg} \mathrm{m}^{-3}$ and its specific heat capacity is $450 \mathrm{Jkg}^{-1} \mathrm{K}^{-1}$ The rate of flow of thermal energy through a surface of area $A$ and thickness $x$ separated by temperatures $T_{1}$ and $T_{2}$ is given by $$\frac{\Delta Q}{\Delta t}=k A \frac{T_{1}-T_{2}}{x}$$ (a) Calculate the mass of the tank. (b) By equating the energy received from the sunlight to the thermal energy lost by conduction to the surrounding air, estimate the final temperature of the water. (c) Find the heat capacity, $C,$ of the tank-water system. (d) Show that the temperature $T$ of the water in $^{\circ} \mathrm{C}$ is increasing at a rate $\frac{\Delta T}{\Delta t}$ that can be found from the equation $C \frac{\Delta T}{\Delta t}=A l_{m}-k A \frac{T-20}{x}$ where $I_{i n}$ is the intensity of sunlight captured by the tank, $C$ is the heat capacity of the system, $k$ is the thermal conductivity of the tank, $A$ is the area of the top surface of the tank, and $x$ is the thickness of the tank wall. (e) Evaluate the rate of temperature increase when the temperature is the average of the initial temperature of $20^{\circ} \mathrm{C}$ and the final temperature you found in part (b). (1) Assuming that the temperature is increasing at this rate, calculate how long it will take the water to reach its final temperature

Sunlight of intensity $800 \mathrm{W} \mathrm{m}^{-2}$ is captured by a tank containing $100 \mathrm{kg}$ of water with an efficiency of $80 \%$. The tank is rectangular in shape and has dimensions $1.0 \times 1.0 \times 0.10 \mathrm{m}^{3}$. It has walls of thickness $5.0 \mathrm{mm} .$ The surrounding air has a temperature of $20^{\circ} \mathrm{C}$. Assume that the tank is well insulated from all sides except the top surface (of area $1.0 \mathrm{m}^{2}$ ). The material of the tank has a thermal conductivity of $k=0.30 \mathrm{Wm}^{-1} \mathrm{K}^{-1},$ its density is
$1200 \mathrm{kg} \mathrm{m}^{-3}$ and its specific heat capacity is $450 \mathrm{Jkg}^{-1} \mathrm{K}^{-1}$
The rate of flow of thermal energy through a surface of area $A$ and thickness $x$ separated by temperatures $T_{1}$ and $T_{2}$ is given by $$\frac{\Delta Q}{\Delta t}=k A \frac{T_{1}-T_{2}}{x}$$ (a) Calculate the mass of the tank.
(b) By equating the energy received from the sunlight to the thermal energy lost by conduction to the surrounding air, estimate the final temperature of the water. (c) Find the heat capacity, $C,$ of the tank-water system.
(d) Show that the temperature $T$ of the water
in $^{\circ} \mathrm{C}$ is increasing at a rate $\frac{\Delta T}{\Delta t}$ that can be found from the equation $C \frac{\Delta T}{\Delta t}=A l_{m}-k A \frac{T-20}{x}$
where $I_{i n}$ is the intensity of sunlight captured by the tank, $C$ is the heat capacity of the system, $k$ is the thermal conductivity of the tank, $A$ is the area of the top surface of the tank, and $x$ is the thickness of the tank wall.
(e) Evaluate the rate of temperature increase when the temperature is the average of the initial temperature of $20^{\circ} \mathrm{C}$ and the final temperature you found in part (b).
(1) Assuming that the temperature is increasing at this rate, calculate how long
it will take the water to reach its final
temperature

Key Concepts

Thermal Conductivity and Fourier's Law

Thermal conductivity is a material property that measures how well a material conducts heat, and Fourier’s Law quantifies this by relating the heat transfer rate to the temperature gradient, the area through which the heat is flowing, and the thickness of the material. Understanding this concept is crucial for analyzing and predicting how heat is lost from or gained by a system through conduction.

Solar Energy Absorption Efficiency

Solar energy absorption involves capturing the sunlight incident on a surface and converting it into thermal energy. The efficiency factor quantifies the fraction of the incident solar energy that is actually absorbed and used in heating the system. This concept is important in evaluating how much energy is available to heat the system from an initial energy source like sunlight.

Heat Transfer Energy Balance

The energy balance in a thermal system involves setting the energy gained (from sources like absorbed solar radiation) equal to the energy lost (through processes like conduction) at equilibrium conditions. This balance helps in determining the steady state or final temperature of the system, ensuring that all energy contributions and losses are appropriately accounted for.

Composite Heat Capacity

Composite heat capacity refers to the total heat capacity of a system that consists of more than one material, such as water and the material of a tank. It is calculated by summing the individual contributions from each component, which is essential for determining the temperature change of the entire system when a given amount of energy is added or removed.

Rate of Temperature Change

The rate of temperature change in a thermal system is derived from the net energy flow into the system and its total heat capacity. By equating the energy input from sources to the energy loss through conduction, a differential equation is established that describes how the temperature evolves over time. This concept is fundamental in predicting transient thermal behavior.

Transcript

00:01 First, we find the volume of the tank.

00:03 So we take the entire volume and then subtract the hollow part.

00:10 So you subtract the thickness of the wall, double it for account for both walls.

00:13 That's two times the point 0 .05 meters, 5 millimeters.

00:17 That will give you the volume in meter cube.

00:19 Then to find the mass of the tank, you take the volume of the tank.

00:24 So that's just the metal content, the metal walls of the tank, multiply by the density of the metal that's 1200 so that will give you a mass of 14 .15 kilogram for the mass of the metal in the wall then for part b we first calculate the power received from the sun so we multiply the intensity with the area exposed then multiply by the efficiency so that will give you 640 watts then we know the temperature of the air is 20 degrees celsius.

00:54 So we'll use that is at a reference.

00:56 Then next step, we set the power received from the sun equal to the power dissipated given by the equation of a question.

01:05 K times a times t1 minus t2 divide by x.

01:09 So t1 is what we are trying to find.

01:11 That's the temperature of the water and t2 is the temperature of the air.

01:15 And then we rearrange the equation and solve for t1.

01:18 So plug in our numbers and we get a temperature of 30 .7 degrees celsius for t1.

01:24 So that's the temperature of the water, this final temperature.

01:28 For part c is given that the mass of the water in the tank is 100 kilogram.

01:33 So to find the heat capacity of the water in the tank, we do this.

01:37 We multiply the mass of the tank with a specific heat of the tank, then the mass of the water with a specific heat of the water.

01:43 So note that in the equations, we are using the specific heat, not just the heat capacity.

01:48 So doing that gives us the heat capacity of the water in the tank together.

01:53 That turns out to be 4 .26 times 10 of the 5 joules per kelvin.

02:05 So this is correction here.

02:07 The unit should be joules per kelvin, not joules per kg kelvin because this is heat capacity, not specific heat.

02:12 Okay, then moving on to part d...

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