The temperature in an electric oven is 160°C. The temperature at the outer surface in the kitchen is 50°C. The oven ( surface area = 1.6 m2) is insulated with material that has a thickness of 0.02 m and a thermal conductivity of 0.045 J/s.m.C°. (a) How much energy is used to operate the oven for six hours?(b) At the price of $0.1/kw-hr for electrical energy, what is the cost of operating the oven?

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To calculate the energy used to operate the oven for six hours, we need to determine the heat transfer through the insulation.

(a) The rate of heat transfer can be calculated using the formula:

Q = k * A * (T1 - T2) / L

where:
Q = rate of heat transfer (in watts)
k = thermal conductivity (in J/s.m.C°)
A = surface area (in square meters)
T1 = temperature on one side of the insulation (in °C)
T2 = temperature on the other side of the insulation (in °C)
L = thickness of the insulation (in meters)

Given:
k = 0.045 J/s.m.C°
A = 1.6 m2
T1 = 160°C
T2 = 50°C
L = 0.02 m

Plugging in the values:

Q = 0.045 * 1.6 * (160 - 50) / 0.02

Q = 0.045 * 1.6 * 110 / 0.02

Q = 0.045 * 1760 / 0.02

Q = 3,960 J/s = 3,960 W (watts)

To calculate the energy used, we multiply the rate of heat transfer by the time:

Energy used = Q * time

Given:
time = 6 hours = 6 * 60 * 60 seconds

Energy used = 3,960 * (6 * 60 * 60)

Energy used = 3,960 * 21,600

Energy used = 85,536,000 J (joules)

(b) Now, let's calculate the cost of operating the oven.

First, we need to convert the energy used from joules to kilowatt-hours (kWh):

1 kWh = 3,600,000 J

Energy used (kWh) = 85,536,000 J / 3,600,000

Energy used (kWh) ≈ 23.76 kWh

Next, we can calculate the cost of operating the oven using the price of $0.1/kWh:

Cost = Energy used (kWh) * Price per kWh

Cost = 23.76 kWh * $0.1/kWh

Cost = $2.376

Therefore, the cost of operating the oven for six hours is approximately $2.376.

To find the energy used to operate the oven for six hours, we need to calculate the heat loss through the insulated walls of the oven. The formula to calculate heat loss through the insulated walls is:

Q = U * A * ΔT * t

Where:
Q = Heat Loss (in Joules)
U = Heat Transfer Coefficient (in J/s.m².C°)
A = Surface Area of the Insulated Walls (in m²)
ΔT = Temperature Difference (in C°)
t = Time (in seconds)

First, we need to convert the time from hours to seconds:

6 hours = 6 * 60 * 60 = 21,600 seconds

Now, let's calculate the temperature difference (ΔT):

ΔT = (T1 - T2)
= (160°C - 50°C)
= 110°C

Next, let's calculate the heat transfer coefficient (U) using the thermal conductivity (k), the thickness of the insulation (L), and the surface area of the walls (A):

U = k / L
= 0.045 J/s.m.C° / 0.02 m
= 2.25 J/m².s.C°

Now, we can substitute the values into the heat loss formula:

Q = U * A * ΔT * t
= 2.25 J/m².s.C° * 1.6 m² * 110°C * 21,600 s
= 839,808 Joules

Therefore, the energy used to operate the oven for six hours is 839,808 Joules.

To find the cost of operating the oven, we need to convert the energy used in Joules to kilowatt-hours (kWh). The conversion is as follows:

1 kilowatt-hour (kWh) = 3.6 x 10^6 Joules

Energy used in kWh = Q / (3.6 x 10^6)

Let's calculate the energy used in kWh:

Energy used in kWh = 839,808 Joules / (3.6 x 10^6)
≈ 0.233 kWh

Now, we can calculate the cost of operating the oven using the price of $0.1/kWh:

Cost = Energy used (kWh) * Price per kWh
= 0.233 kWh * $0.1/kWh
= $0.0233

Therefore, the cost of operating the oven is approximately $0.0233.

Calculate the heat loss rate through the walls, using

P = (surface area)*k*(110C)/0.02 m
where k is the thermal conducticity

Convert watts (J/s) to kilowatts. Multiply the number of kW by 6h to get the number of kW-hr.

That is the energy required for steady state ooperation for 6 hours. There will be an additional amount required to heat up the oven - probably a few percent more, depending upon the specific heat.