Question

Question 1: A rock of mass 0.453 kg falls from rest from a height of 23.0 m into a pail containing 0.335 kg of water. The rock and water have the same initial temperature. The specific heat capacity of the rock is 1920 J/kg C°. Ignore the heat absorbed by the pail itself, and determine the rise in temperature of the rock and water in Celsius degrees.

Question 2: A 33.7-kg block of ice at 0 °C is sliding on a horizontal surface. The initial speed of the ice is 7.22 m/s and the final speed is 2.93 m/s. Assume that the part of the block that melts has a very small mass and that all the heat generated by kinetic friction goes into the block of ice, and determine the mass of ice that melts into water at 0 °C.

Answer 1

By the energy conservation law:

The potential energy of the falling rock is ultimately converted into heat:

mgh = Q = mCpΔT + m' Cp' ΔT

where,

mass of rock: m = 0.453 kg

mass of water: m' = 0.335 kg

specific heat of rock: Cp = 1845 J/kg-C

specific heat of water: Cp' = 4186 J/kg-C

Temperature rise: ΔT

mgh = Q = mCpΔT + m' Cp' ΔT

mgh = (mCp + m' Cp') ΔT

0.453 * 9.81 * 23 = (0.453*1845 + 0.335*4186) ΔT

102.21 = (835.78 + 1402.31) ΔT

(102.21 / 2238.09 ) = ΔT

ΔT = 0.0456 ^{o C}