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

I hope you see this response, as it looks like something happened and you have 5+ copies of this post in the forum.

So this problem just deals with heat exchange and energy conservation. The premise is that the water gains a bunch of energy as it rises to 26.8 degrees, so we are meant to assume (since it's "insulated") that all of the energy the water gains has come from the lead ball, which must have been at a temperature higher than 26.8 degrees to begin with.

You should have a formula for heat energy gained or lost. It's usually written as Q = m*c*ΔT, where Q is the heat energy gained or lost, m is mass, c is the specific heat, and ΔT is the temperature change. So, your Q value for the water will be equal but opposite to the Q lost by the lead ball. Solve for ΔT to find how much its temp changed.

When you look up your "c" value, just make sure the units involve the same ones you're using - kg and degrees Celsius. Oh and you'll need to figure out how much mass 2.5L: of water is. That's easy enough to google.