A brass boiler has a base area of 0.15 m^{2} 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. Thermal conductivity of brass = 109 J s^{–1} m^{–1 }K^{–1}; Heat of vaporisation of water = 2256 × 10^{3} J kg^{–1}.

Asked by Abhisek | 1 year ago | 201

Base area of the brass boiler, A= 0.15 m^{2}

Thickness of the boiler, d =1.0 cm=0.01 m

Brass boiler boils water at the rate, R=6.0 kg/min

Time, t = 60 sec

Mass = 6 kg

Thermal conductivity of brass, K =109 Js^{−1}m^{−1}k^{−1}

Heat of vaporisation of water, L = 2256×10^{3} Jkg^{−1}

Let T_{1} Temperature of the flame in contact with the boiler

Let T_{2} be the boiling point of water = 100°C

The Q be the amount of heat flowing into water through the base of the boiler

Q =\( \dfrac{KA(T_1 −T_2)t}{d}\)

Q = \(\dfrac{109 \times 0.15 \times (T_1 – 100) \times 60}{0.01}\) ..........(1)

Heat received by water:

Q=mL

Q = 6 x 2256×10^{3} ——-(2)

Equating equations (1) and (2), we get:

mL = \( \dfrac{KA(T_1 −T_2)t}{d}\)

6 x 2256×10^{3}

= \( \dfrac{ [10^9 \times 0.15 \times (T_1 – 100) \times 60]}{ 0.01}\)

13536 ×10^{3} x 0.01 = 981 (T_{1} – 100)

T_{1} – 100 = \( \dfrac{ 13536 ×10^3 \times 0.01}{ 981}\)

T_{1} = 137.9 + 100

= 237. 9^{°}C

Therefore, the temperature of the part of the flame in contact with the boiler is 237.98°C.

Answered by Pragya Singh | 1 year agoA hot ball cools from 90°C to 10°C in 5 minutes. If the surrounding temperature is 20°C, what is the time taken to cool from 60°C to 30°C?

Answer the following questions based on the P-T phase diagram of carbon dioxide:

**(a)** At what temperature and pressure can the solid, liquid and vapour phases of \( CO_2\) co-exist in equilibrium?

**(b)** What is the effect of the decrease of pressure on the fusion and boiling point of \( CO_2\)?

**(c) **What are the critical temperature and pressure for \( CO_2\)? What is its significance?

**(d)** Is \( CO_2\) solid, liquid or gas at

**(a) **–70°C under 1 atm,

**(b)** –60°C under 10 atm,

**(c) **15°C under 56 atm?

A body cools from 80°C to 50°C in 5 minutes. Calculate the time it takes to cool from 60°C to 30°C. The temperature of the surroundings is 20°C.

Explain why :

**(a)** a body with large reflectivity is a poor emitter

**(b)** a brass tumbler feels much colder than a wooden tray on a chilly day

**(c)** an optical pyrometer (for measuring high temperatures) calibrated for an ideal black body radiation gives too low a value for the temperature of a red hot iron piece in the open, but gives a correct value for the temperature when the same piece is in the furnace

**(d) **the earth without its atmosphere would be inhospitably cold

**(e)** heating systems based on the circulation of steam are more efficient in warming a building than those based on the circulation of hot water

A ‘thermacole’ icebox is a cheap and an 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 the co-efficient 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]