Advertisements
Advertisements
Question
A ‘thermacole’ icebox is a cheap and efficient method for storing small quantities of cooked food in summer in particular. A cubical icebox of side 30 cm has a thickness of 5.0 cm. If 4.0 kg of ice is put in the box, estimate the amount of ice remaining after 6 h. The outside temperature is 45 °C, and coefficient of thermal conductivity of thermacole is 0.01 J s–1 m–1 K–1. [Heat of fusion of water = 335 × 103 J kg–1]
Solution 1
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 × 60 × 60 s
Outside temperature, T = 45°C
Coefficient of thermal conductivity of thermacole, K = 0.01 J s–1 m–1 K–1
Heat of fusion of water, L = 335 × 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: `theta = (KA(T-0)t)/l`
Where
A = Surface area of the box = 6s2 = 6 × (0.3)2 = 0.54 m3
`theta = (0.01xx0.54xx(45)xx6xx60xx60)/0.05 = 104976 J`
But `theta = m'L`
`:. m' = theta/L`
`= 104976/(335xx10^3) = 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.
Solution 2
Each side of the cubical box (having 6 faces) is 30 cm = 0.30 m. Therefore, the total surface area’ of the icebox exposed to outside air is A = 6 x (0.30 m)2 = 0.54 m2. The thickness of the icebox is d = 5.0 cm = 0.05 m, time of exposure t = 6h = 6 x 3600 s and temperature difference T1 – T2 = 45°C – 0°C = 45°C.
∴Total heat entering the icebox in 6 h is given by
`Q = (KA(T1 - T2)t)/d`
`= (0.01 Js^(-1) m^(-1) C6(-1)xx0.54 m^2 xx 45 ^@C xx (6xx 3600 s))/"0.05m"`
`= 1.05 xx10^5 J`
Suppose a mass m of ice melts with this heat. Then Q = mL, where L is tatent heat of fusion of water thus
`1.05 xx 10^5 J = m(335xx10^3) "JKg"^(-1)`
or `m = (1.05 xx10^5 J)/(335 xx 10^3 J kg^(-1)) = 0.313 kg`
The initial mass of ice in the box is 4.0 kg. Therefore the ice remaining in the box after 6 h is = (4.0 - 0.313) kg
= 3.687 kg
Solution 3
Each side of the cubical box (having 6 faces) is 30 cm = 0.30 m. Therefore, the total surface area’ of the icebox exposed to outside air is A = 6 x (0.30 m)2 = 0.54 m2. The thickness of the icebox is d = 5.0 cm = 0.05 m, time of exposure t = 6h = 6 x 3600 s and temperature difference T1 – T2 = 45°C – 0°C = 45°C.
∴Total heat entering the icebox in 6 h is given by
`Q = (KA(T1 - T2)t)/d`
`= (0.01 Js^(-1) m^(-1) C6(-1)xx0.54 m^2 xx 45 ^@C xx (6xx 3600 s))/"0.05m"`
`= 1.05 xx10^5 J`
Suppose a mass m of ice melts with this heat. Then Q = mL, where L is tatent heat of fusion of water thus
`1.05 xx 10^5 J = m(335xx10^3) "JKg"^(-1)`
or `m = (1.05 xx10^5 J)/(335 xx 10^3 J kg^(-1)) = 0.313 kg`
The initial mass of ice in the box is 4.0 kg. Therefore the ice remaining in the box after 6 h is = (4.0 - 0.313) kg
= 3.687 kg
APPEARS IN
RELATED QUESTIONS
Answer the following question based on the P-T phase diagram of carbon dioxide:
Is CO2 solid, liquid or gas at
- –70 °C under 1 atm,
- –60 °C under 10 atm,
- 15 °C under 56 atm?
Answer the following questions based on the P–T phase diagram of CO2:
What happens when CO2 at 4 atm pressure is cooled from room temperature at constant pressure?
Answer the following questions based on the P–T phase diagram of CO2:
CO2 is heated to a temperature 70 °C and compressed isothermally. What changes in its properties do you expect to observe?
A brass boiler has a base area of 0.15 m2 and thickness 1.0 cm. It boils water at the rate of 6.0 kg/min when placed on a gas stove. Estimate the temperature of the part of the flame in contact with the boiler. The thermal conductivity of brass = 109 J s –1 m–1 K–1; Heat of vaporisation of water = 2256 × 103 J kg–1.
A metal block of heat capacity 80 J°C−1 placed in a room at 20°C is heated electrically. The heater is switched off when the temperature reaches 30°C. The temperature of the block rises at the rate of 2°C s−1 just after the heater is switched on and falls at the rate of 0.2°C s−1 just after the heater is switched off. Assume Newton's law of cooling to hold.
- Find the power of the heater.
- Find the power radiated by the block just after the heater is switched off.
- Find the power radiated by the block when the temperature of the block is 25°C.
- Assuming that the power radiated at 25°C represents the average value in the heating process, find the time for which the heater was kept on.
Answer the following question based on the P-T phase diagram of carbon dioxide:
At what temperature and pressure can the solid, liquid and vapour phases of CO2 co-exist in equilibrium?
Answer the following question based on the P-T phase diagram of carbon dioxide:
What are the critical temperature and pressure for CO2? What is their significance?
Mark the correct options:
- A system X is in thermal equilibrium with Y but not with Z. System Y and Z may be in thermal equilibrium with each other.
- A system X is in thermal equilibrium with Y but not with Z. Systems Y and Z are not in thermal equilibrium with each other.
- A system X is neither in thermal equilibrium with Y nor with Z. The systems Y and Z must be in thermal equilibrium with each other.
- A system X is neither in thermal equilibrium with Y nor with Z. The system Y and Z may be in thermal equilibrium with each other.
During summers in India, one of the common practice to keep cool is to make ice balls of crushed ice, dip it in flavoured sugar syrup and sip it. For this a stick is inserted into crushed ice and is squeezed in the palm to make it into the ball. Equivalently in winter, in those areas where it snows, people make snow balls and throw around. Explain the formation of ball out of crushed ice or snow in the light of P–T diagram of water.
100 g of water is supercooled to –10°C. At this point, due to some disturbance mechanised or otherwise some of it suddenly freezes to ice. What will be the temperature of the resultant mixture and how much mass would freeze?
`[S_w = 1cal/g/^circC and L_(Fusion)^w = 80cal/g]`
