A rock of mass 0.221 kg falls from rest from a height of 27.2 m Into a pail cont

Question

A rock of mass 0.221 kg falls from rest from a height of 27.2 m Into a pail containing 0.334 kg of water. The rock and water have the same initial temperature. The specific heat capacity of the rock is 1930 J/kg C degree. Ignore the heat absorbed by the pail itself, and determine the rise in temperature of the rock and water In Celsius degrees. Number Units the tolerance is +/-2% By accessing this Question Assistance, you will learn while you earn points based on the Point Potential Policy set by your instructor.

Explanation / Answer

the potential energy of the rock will be converted into heat energy gained by both rock and the water.

let rise in temperature is T degree celcius.

then using energy conservation principle:

initial potential energy of the rock=heat energy gained by rock + heat energy gained by water

==>mass of rock*g*initial height=mass of rock*specific heat*temperature change+mass of water*specific heat of water*temperature change

==>0.221*27.2*9.8=0.221*1930*T+0.334*4186*T

==>T=0.221*27.2*9.8/(0.221*1930+0.334*4186)=0.03228 degree celcius

so temperature rise is 0.03228 degree celcius

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