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A thermocol cubical ice box of side 30cm has a thickness of 5cm. If 4kg of ice is put in the box. Then estimate the amount of ice remained after 6hours.the outside temperature is 45 degree celcius , the coefficient of thermal conductivity of the thermocol is 0.01 watt / m degree celcius and the heat of fusion of water is 335 ×10^3 J/kg ? the answer for this is 3.687kg.?
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A thermocol cubical ice box of side 30cm has a thickness of 5cm. If 4k...
Side of the given cubical ice box, s = 30 cm = 0.3 m
Thickness of the ice box, l = 5.0 cm = 0.05 m
Mass of ice kept in the ice box, m = 4 kg
Time gap, t = 6 h = 6 x 60 x 60 s
Outside temperature, T = 45degC
Coefficient of thermal conductivity of thermacole, K = 0.01 J s–1 m–1 K–1
Heat of fusion of water, L = 335 x 103 J kg–1
Let m’ be the total amount of ice that melts in 6 h.
The amount of heat lost by the food:
θ = KA(T – 0)t / l
Where,
A = Surface area of the box = 6s2 = 6 x (0.3)2 = 0.54 m3
θ = 0.01 x 0.54 x 45 x 6 x 60 x 60 / 0.05  =  104976 J
But θ = m’L
∴ m‘ = θ/L
= 104976/(335 x 103)  =  0.313 kg
Mass of ice left = 4 – 0.313 = 3.687 kg
Hence, the amount of ice remaining after 6 h is 3.687 kg.
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A thermocol cubical ice box of side 30cm has a thickness of 5cm. If 4k...
Given:
- Side of the ice box (L) = 30 cm
- Thickness of the ice box (d) = 5 cm
- Mass of ice (m) = 4 kg
- Time (t) = 6 hours = 6 × 3600 seconds
- Outside temperature (T1) = 45 degrees Celsius
- Coefficient of thermal conductivity of thermocol (k) = 0.01 watt/m degree Celsius
- Heat of fusion of water (H) = 335 × 10^3 J/kg

To Find:
The amount of ice remained after 6 hours.

Assumptions:
- The ice box is well-insulated, except for the top surface.
- The ice box is initially at a uniform temperature.
- The rate of heat transfer through the sides and bottom of the ice box is negligible compared to the top surface.
- The temperature of the ice box remains constant during the melting process.

Solution:

Step 1: Calculate the surface area of the top surface of the ice box.

The surface area of a cube is given by:
A = 6 × (side)^2

Given that the side of the ice box is 30 cm, the surface area of the top surface is:
A = 6 × (30 cm)^2

Step 2: Calculate the rate of heat transfer through the top surface of the ice box.

The rate of heat transfer through a material is given by:
Q = k × A × (T1 - T2) / d

Where:
Q = Rate of heat transfer (in watts)
k = Coefficient of thermal conductivity of the material (in watt/m degree Celsius)
A = Surface area of heat transfer (in m^2)
T1 = Outside temperature (in degrees Celsius)
T2 = Temperature inside the ice box (in degrees Celsius)
d = Thickness of the material (in meters)

Since the temperature inside the ice box remains constant at 0 degrees Celsius during the melting process, T2 = 0 degrees Celsius.

Substituting the given values, we have:
Q = 0.01 × A × (45 - 0) / d

Step 3: Calculate the total heat transferred to melt the ice.

The total heat transferred is given by:
Q_total = Q × t

Substituting the given values, we have:
Q_total = (0.01 × A × (45 - 0) / d) × (6 × 3600)

Step 4: Calculate the mass of ice melted.

The heat required to melt a certain amount of ice is given by:
Q_melt = m × H

Where:
Q_melt = Heat required to melt the ice (in joules)
m = Mass of ice (in kg)
H = Heat of fusion of water (in J/kg)

Substituting the given values, we have:
Q_melt = 4 kg × (335 × 10^3 J/kg)

Step 5: Calculate the amount of ice remained.

The amount of ice remained can be calculated using the equation:
m_remained = m - (Q_total / Q_melt)

Substituting the calculated values, we have:
m_remained = 4 kg - (
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A thermocol cubical ice box of side 30cm has a thickness of 5cm. If 4kg of ice is put in the box. Then estimate the amount of ice remained after 6hours.the outside temperature is 45 degree celcius , the coefficient of thermal conductivity of the thermocol is 0.01 watt / m degree celcius and the heat of fusion of water is 335 ×10^3 J/kg ? the answer for this is 3.687kg.?
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A thermocol cubical ice box of side 30cm has a thickness of 5cm. If 4kg of ice is put in the box. Then estimate the amount of ice remained after 6hours.the outside temperature is 45 degree celcius , the coefficient of thermal conductivity of the thermocol is 0.01 watt / m degree celcius and the heat of fusion of water is 335 ×10^3 J/kg ? the answer for this is 3.687kg.? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A thermocol cubical ice box of side 30cm has a thickness of 5cm. If 4kg of ice is put in the box. Then estimate the amount of ice remained after 6hours.the outside temperature is 45 degree celcius , the coefficient of thermal conductivity of the thermocol is 0.01 watt / m degree celcius and the heat of fusion of water is 335 ×10^3 J/kg ? the answer for this is 3.687kg.? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A thermocol cubical ice box of side 30cm has a thickness of 5cm. If 4kg of ice is put in the box. Then estimate the amount of ice remained after 6hours.the outside temperature is 45 degree celcius , the coefficient of thermal conductivity of the thermocol is 0.01 watt / m degree celcius and the heat of fusion of water is 335 ×10^3 J/kg ? the answer for this is 3.687kg.?.
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