Question

A piece of iron mass equals 25.0 g at 398 Calvin is placed in a Styrofoam cup containing 25.0 mL of water at 298 K assuming that no heat is lost to the cup or the surroundings what will the final temperature of the water

Answer 1

Final Temperature = 308K

We can set this up as

-Q1 = Q2

Q1 = (iron mass)•(specific heat of iron)•(T2-T1)

[where T2 means final temperature and T1 means initial temperature of the iron]

Q2 = (water mass)•(specific heat of water)•(T2-T3)

[where T3 means inital temperature of the water]

Plug in what you know

Q1 = (25.0g)•(0.44J/g•K)•(T2-398K)

[The internet can tell us the specific heat of iron, and the temperature unit is Kelvin]

Q2 = (25.0g)•(4.18J/g•K)•(T2-298K)

[The internet can tell us the specific heat of water, and 1mL = 1g]

Multiply them all

Q1 = 11(T2) - 4378K

Q2 = 104.5(T2) - 31141K

Use these values in the first equation and solve

-Q1 = Q2

-(11(T2) - 4378K) = 104.5(T2) - 31141K

-11(T2) + 4378K = 104.5(T2) - 31141K

35519K = 115.5(T2)

T2 = 308K

[This is not exact since there was some rounding; but it will be very close]