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A cold storage compartment is 4.5 m long by 4 m wide by 2.5 high. The four walls, ceiling and floor are covered to a thickness of 150 mm with insulating material which has coefficient of thermal conductivity of 0.058 W/m K . calculate the quantity of heat leaking through insulation per hour when outside face temperatures of material is 15 degree Centigrade and -5 degree Centigrade. solve using heat transfer method and put the answer in degree centigrade and joules per second?
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A cold storage compartment is 4.5 m long by 4 m wide by 2.5 high. The ...
**Problem Analysis:**
In this problem, we need to calculate the quantity of heat leaking through the insulation of a cold storage compartment. To solve this problem, we can use the heat transfer method and apply the equation for heat conduction through a material.
**Given Data:**
- Length of the cold storage compartment (L) = 4.5 m
- Width of the cold storage compartment (W) = 4 m
- Height of the cold storage compartment (H) = 2.5 m
- Thickness of the insulation material (t) = 150 mm = 0.15 m
- Coefficient of thermal conductivity of the insulation material (k) = 0.058 W/m K
- Outside face temperature of the material (T1) = 15 °C
- Inside face temperature of the material (T2) = -5 °C
**Solution:**
To calculate the quantity of heat leaking through the insulation, we can use the equation for heat conduction:
Q = (k * A * (T1 - T2)) / d
where:
- Q is the quantity of heat leaking through the insulation (in joules per second or watts)
- k is the coefficient of thermal conductivity of the insulation material (in watts per meter kelvin)
- A is the surface area of the insulation material (in square meters)
- T1 is the outside face temperature of the material (in kelvin)
- T2 is the inside face temperature of the material (in kelvin)
- d is the thickness of the insulation material (in meters)
**Step 1: Calculate the Surface Area of the Insulation Material:**
The surface area of the insulation material can be calculated by finding the areas of the six sides of the cold storage compartment.
- Area of the four walls = 2 * (L * H + W * H)
- Area of the ceiling = L * W
- Area of the floor = L * W
Total surface area (A) = Area of the four walls + Area of the ceiling + Area of the floor
**Step 2: Convert Temperatures to Kelvin:**
To use the temperature values in the equation, we need to convert them from degrees Celsius to kelvin.
T1 (in kelvin) = T1 (in °C) + 273
T2 (in kelvin) = T2 (in °C) + 273
**Step 3: Calculate the Quantity of Heat Leaking Through the Insulation:**
Using the given values and the calculated values, we can now substitute them into the equation for heat conduction:
Q = (k * A * (T1 - T2)) / d
Substitute the values and solve for Q.
**Step 4: Convert the Quantity of Heat to Joules per Second:**
The quantity of heat is given in joules per second or watts. So, the calculated value of Q represents the quantity of heat leaking through the insulation per second.
**Step 5: Convert the Quantity of Heat to Joules per Hour:**
To convert the quantity of heat from joules per second to joules per hour, we can multiply the value of Q by 3600 (since there are 3600 seconds in an hour).
**Step 6: Present the Final Answer:**
The final answer will be in the form of "X degrees Celsius and Y joules per
In this problem, we need to calculate the quantity of heat leaking through the insulation of a cold storage compartment. To solve this problem, we can use the heat transfer method and apply the equation for heat conduction through a material.
**Given Data:**
- Length of the cold storage compartment (L) = 4.5 m
- Width of the cold storage compartment (W) = 4 m
- Height of the cold storage compartment (H) = 2.5 m
- Thickness of the insulation material (t) = 150 mm = 0.15 m
- Coefficient of thermal conductivity of the insulation material (k) = 0.058 W/m K
- Outside face temperature of the material (T1) = 15 °C
- Inside face temperature of the material (T2) = -5 °C
**Solution:**
To calculate the quantity of heat leaking through the insulation, we can use the equation for heat conduction:
Q = (k * A * (T1 - T2)) / d
where:
- Q is the quantity of heat leaking through the insulation (in joules per second or watts)
- k is the coefficient of thermal conductivity of the insulation material (in watts per meter kelvin)
- A is the surface area of the insulation material (in square meters)
- T1 is the outside face temperature of the material (in kelvin)
- T2 is the inside face temperature of the material (in kelvin)
- d is the thickness of the insulation material (in meters)
**Step 1: Calculate the Surface Area of the Insulation Material:**
The surface area of the insulation material can be calculated by finding the areas of the six sides of the cold storage compartment.
- Area of the four walls = 2 * (L * H + W * H)
- Area of the ceiling = L * W
- Area of the floor = L * W
Total surface area (A) = Area of the four walls + Area of the ceiling + Area of the floor
**Step 2: Convert Temperatures to Kelvin:**
To use the temperature values in the equation, we need to convert them from degrees Celsius to kelvin.
T1 (in kelvin) = T1 (in °C) + 273
T2 (in kelvin) = T2 (in °C) + 273
**Step 3: Calculate the Quantity of Heat Leaking Through the Insulation:**
Using the given values and the calculated values, we can now substitute them into the equation for heat conduction:
Q = (k * A * (T1 - T2)) / d
Substitute the values and solve for Q.
**Step 4: Convert the Quantity of Heat to Joules per Second:**
The quantity of heat is given in joules per second or watts. So, the calculated value of Q represents the quantity of heat leaking through the insulation per second.
**Step 5: Convert the Quantity of Heat to Joules per Hour:**
To convert the quantity of heat from joules per second to joules per hour, we can multiply the value of Q by 3600 (since there are 3600 seconds in an hour).
**Step 6: Present the Final Answer:**
The final answer will be in the form of "X degrees Celsius and Y joules per
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A cold storage compartment is 4.5 m long by 4 m wide by 2.5 high. The four walls, ceiling and floor are covered to a thickness of 150 mm with insulating material which has coefficient of thermal conductivity of 0.058 W/m K . calculate the quantity of heat leaking through insulation per hour when outside face temperatures of material is 15 degree Centigrade and -5 degree Centigrade. solve using heat transfer method and put the answer in degree centigrade and joules per second?
Question Description
A cold storage compartment is 4.5 m long by 4 m wide by 2.5 high. The four walls, ceiling and floor are covered to a thickness of 150 mm with insulating material which has coefficient of thermal conductivity of 0.058 W/m K . calculate the quantity of heat leaking through insulation per hour when outside face temperatures of material is 15 degree Centigrade and -5 degree Centigrade. solve using heat transfer method and put the answer in degree centigrade and joules per second? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A cold storage compartment is 4.5 m long by 4 m wide by 2.5 high. The four walls, ceiling and floor are covered to a thickness of 150 mm with insulating material which has coefficient of thermal conductivity of 0.058 W/m K . calculate the quantity of heat leaking through insulation per hour when outside face temperatures of material is 15 degree Centigrade and -5 degree Centigrade. solve using heat transfer method and put the answer in degree centigrade and joules per second? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A cold storage compartment is 4.5 m long by 4 m wide by 2.5 high. The four walls, ceiling and floor are covered to a thickness of 150 mm with insulating material which has coefficient of thermal conductivity of 0.058 W/m K . calculate the quantity of heat leaking through insulation per hour when outside face temperatures of material is 15 degree Centigrade and -5 degree Centigrade. solve using heat transfer method and put the answer in degree centigrade and joules per second?.
A cold storage compartment is 4.5 m long by 4 m wide by 2.5 high. The four walls, ceiling and floor are covered to a thickness of 150 mm with insulating material which has coefficient of thermal conductivity of 0.058 W/m K . calculate the quantity of heat leaking through insulation per hour when outside face temperatures of material is 15 degree Centigrade and -5 degree Centigrade. solve using heat transfer method and put the answer in degree centigrade and joules per second? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A cold storage compartment is 4.5 m long by 4 m wide by 2.5 high. The four walls, ceiling and floor are covered to a thickness of 150 mm with insulating material which has coefficient of thermal conductivity of 0.058 W/m K . calculate the quantity of heat leaking through insulation per hour when outside face temperatures of material is 15 degree Centigrade and -5 degree Centigrade. solve using heat transfer method and put the answer in degree centigrade and joules per second? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A cold storage compartment is 4.5 m long by 4 m wide by 2.5 high. The four walls, ceiling and floor are covered to a thickness of 150 mm with insulating material which has coefficient of thermal conductivity of 0.058 W/m K . calculate the quantity of heat leaking through insulation per hour when outside face temperatures of material is 15 degree Centigrade and -5 degree Centigrade. solve using heat transfer method and put the answer in degree centigrade and joules per second?.
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