A 3.8 kg lead ball is dropped into a 2.5 L insulated pail of water initially at 20.0∘C.

Question

A 3.8 kgkg lead ball is dropped into a 2.5 LL insulated pail of water initially at 20.0∘C∘C.If the final temperature of the water-lead combination is 26.8 ∘C∘C, what was the initial temperature of the lead ball?

Express your answer using two significant figures.

Answer 1

Using conservation of heat energy

Heat given out by Lead ball = Heat taken by water

Heat given by ball

Mass of Ball * Sp heat pf Lead* Fall in temperature

Heat taken by water = Mass of water* Sp heat of Water*Rise in Temp of water

Plug in values

3.8*159 * (Theta-26.8) = 2.5*4200*( 26.8-20.0)

Get value of Theta