Question

I need help with this Chemistry problem, how do I do this?

Answer 1

See below

Explanation:

662C is pretty hot, but I'm going to assume the water didn't sizzle (vaporize) when the metal was placed into the water - this would mess up the calculations a bit.

Metal cools - and loses a bunch of heat.
`Q = cmΔT` This the basic energy change equation that uses specific heat
`Q = 0.897J/"gC"xx3.0gxx(90.3-662)` = -1583.4 J

This 1583.4 J came out of the metal (that is why it is negative)

This 1583.4 J will go into the water to heat it up (positive)
`Q = cmΔT`
we'll solve for m...mass c for water is 4.184 J/gC
`m = Q/"cΔT"`
`m = "1583.4J"/"4.184J/gC"(90.3-15)"`
m = 5.03 grams water

volume = 5.03g(1mL/g) = 5.03 mL water