A 45.2 g sample of Silver at 101.0 ºC is dropped into 300.0 g of water at 30.6 ºC. What is the final temperature of the water?

Final temperature of water = 31.2° C

Explanation:

Let the final temperature of water be T degrees

Concept: By energy conservation law

Heat lost by the silver = heat gained by water

Heat lost be silver Q1 = m1c1ΔT1

Q1 = 45.2/1000Kg×2.36J/KgK×(101-T)

Heat gained by water Q2 = m2c2ΔT2

Q2 = 300/1000Kg×4.186J/KgK(T-30.6)

Since, Q1=Q2

⇒  45.2/1000Kg×2.36J/KgK×(101-T) = 300/1000Kg×4.186J/KgK(T-30.6)

Solving we get T = 31.2° C

Therefore, final temperature of water = 31.2° C

ca has a crystalline lattice with positive ions (since it is a metal, metals are positively charged) and electrons flow freely in its structure since the ca atoms have a large nuclear radius so the electrons are not tightly held, therefore the electrons can flow freely since they are not being highly attracted by the nucleus. moreover, one of the properties of metals are that they are conductors of electricity, thus proving ca is the element.