Question

A lead ball, with an initial temperature of 25 °C, is released from a height of 125.0 m. It does not bounce when it hits a hard surface. Assume all the energy of the fall goes into heating the lead. Find the temperature in °C of the ball after it hits. (You do not need to enter the units.) Data: c_lead = 128 [(J)/(kg·^°C)].

Answer 1

Given data :

Initial temperature of lead ball = 25°C

Ball is dropped from height = 125 m

Specific heat capacity of lead ball = 128 J/kg °C

Concept:

Pottential energy of ball will be converted in to kinetic energy during falling and then this energy will be converted into heat.

It mention to assume all energy goes to heat the ball,

Let m = mass of ball, (kg)

Pottential energy = mgH (J)

From energy conservation equation

or        

T = Final temperature of lead ball

T = 34.58                   ( Unit will be in °C )