Answer:
Explanation:
We shall solve this problem on the basis of averaging , otherwise it will require integration which is a complex operation .
According to newton's law of cooling
dQ / dt = k ( T₁ - T₂ )
T₁ is temperature of surrounding and T₂ is temperature of object .
For the heating by 2 degree
dQ = ms x ΔT , m is mass , s is specific heat and ΔT is rise of
= ms x 2
dt = 1 second
T₁ the average temperature of object = (30 + 32) / 2 = 31
dQ = ms x ΔT
ms x 2 = k ( 100 - 31 )
k = 2 ms / 69
In the second case bar's temperature rises from 32 to 70
average temperature = 32 + 70 / 2 = 51
If t be the time required
dQ = ms x ( 70 - 32 ) = 38ms
38ms / t = k ( 100 - 51 )
38ms / t = (2ms / 69) x 49
t = 38 x 69 / (2 x 49)
= 26.75 s .
