Question

An iron ball falls from a height of 1 km of the ground. If all the energy is converted into heat, the rise in temperature of the ball will be (sp. heat of iron=0.1)

a). 23.3 degree celcius.                      B). 2.33 degree celcius
c). 233 degree celcius.                       D). 0.233 degree celcius

Answer 1

let mass of iron ball = m
At height = 1km =1000m, 
total energy = potential energy = mgh = 1000*9.8*m = 9800m J =9.8m kJ
When all energy is converted into heat and used in raising the temperature of ball,
Specific heat of iron(s) = 0.1 kcal/kg C = 0.42 kJ/kg C 
Let change in temperature = ΔT
ms(ΔT) = mgh
⇒ms(ΔT) = 9.8m
⇒ s(ΔT) = 9.8
⇒ 0.42(ΔT) = 9.8
⇒ ΔT = 9.8/0.42 = 23.3 degree celsius

Answer 2

let mass of iron ball = m

At height = 1km =1000m, 

total energy = potential energy = mgh = 1000*9.8*m = 9800m J =9.8m kJ

When all energy is converted into heat and used in raising the temperature of ball,

Specific heat of iron(s) = 0.1 kcal/kg C = 0.42 kJ/kg C 

Let change in temperature = ΔT

ms(ΔT) = mgh

⇒ms(ΔT) = 9.8m

⇒ s(ΔT) = 9.8

⇒ 0.42(ΔT) = 9.8

⇒ ΔT = 9.8/0.42 = 23.3 degree celsius