Question

two sphere of radii 1:2and densities in the ratio 2:1 and of same specific heat , are heated to same temprature and left in same surrounding . their rate of falling temprature will be in the ratio.​

Answer 1

Answer:

Given-

r1/r2 = 1/2

d1/d2 = 2/1

# Solution-

Rate of cooling is given by -

dθ/dt = (m × c) / (A × T)

dθ/dt = (V×d × c) / (A × T)

dθ/dt = (4/3)πr^3d × c / (πr^2 × T)

dθ/dt = (4/3)rdc / T

For both spheres, c & T are constant, therefore-

(dθ1/dt) / (dθ2/dt) = (4r1d1c/3T) / (4r2d2c/3T)

(dθ1/dt) / (dθ2/dt) = r1d1 / r2d2

(dθ1/dt) / (dθ2/dt) = (r1/r2)(d1/d2)

(dθ1/dt) / (dθ2/dt) = 1/2 × 2

(dθ1/dt) / (dθ2/dt) = 1

Ratio of temp. cooling for spheres is 1:1 .

Explanation:

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Answer 2

Answer:

Hey buddy,

● Answer-

1:1

● Explaination-

# Given-

r1/r2 = 1/2

d1/d2 = 2/1

# Solution-

Rate of cooling is given by -

dθ/dt = (m × c) / (A × T)

dθ/dt = (V×d × c) / (A × T)

dθ/dt = (4/3)πr^3d × c / (πr^2 × T)

dθ/dt = (4/3)rdc / T

For both spheres, c & T are constant, therefore-

(dθ1/dt) / (dθ2/dt) = (4r1d1c/3T) / (4r2d2c/3T)

(dθ1/dt) / (dθ2/dt) = r1d1 / r2d2

(dθ1/dt) / (dθ2/dt) = (r1/r2)(d1/d2)

(dθ1/dt) / (dθ2/dt) = 1/2 × 2

(dθ1/dt) / (dθ2/dt) = 1

Ratio of temp. cooling for spheres is 1:1 .

Hope this helps..