Question

A cubical vassel of side 50mm of negligible heat capacity is filled with water. It is placed in room whose walls are maintained at a constant temperature of 26.00C. According to Newton’s law of cooling; the rate of heat lost, dQ/dt is

dQ/dt = KA (θ - θ₀)

where A is area of cross-section of body losing heat θ – θ₀ = excess temperature of body over surroundings k = 5.00 x 10¹ Cal s^-1 m^-2 (°C)^-1. The time taken by water in vassel to cool from 30°C to 28°C is

(1) 10³ s

(2) 1111 s

(3) 911 s

(4) 1200 s

Answer 1

(2) 1111 s

M = Mass of water in cubical vassel

= (5 x 10^-2)³ x 10³ kg = 125 x 10^-3 kg

H = The heat lost by water in vassel

= 125 x 10^-3 x 1 x 2 k cal = 250 x 10^-3 k cal

Let t be the time taken. The average rate of cooling

The surface area of cubical vassel = 6 x [5 x 10^-2]² m²

= 150 x 10^-4 m²

The average excess temperature of water in vassel over surroundings = (30+28/2) - 26.0 = 3°C. from

Newton’s law

R = 5.00 x 10¹ x 150 x 10^-4 x 3 cal s^-1 

= 225 x 10^-3 cal s^-1 .....(2)

From Eqns. (1) and (2) we have