Question

A hot lump of 37.8 g of iron at an initial temperature of 51.3 °C is placed in 50.0 mL of H2O initially at 25.0 °C and allowed to reach thermal equilibrium. What is the final temperature of the iron and water given that the specific heat of iron is 0.449 J/(g·°C)? Assume no heat is lost to surroundings.

Answer 1

Let us denote water by symbol 1 and iron by symbol 2
volume of water is 50.0 mL and density is 1 g/mL. So,
mass of water = density * volume
= 1 g/mL * 50.0 mL
= 50.0 g
m1 = 50.0 g
T1 = 25.0 oC
C1 = 4.184 J/goC
m2 = 37.8 g
T2 = 51.3 oC
C2 = 0.449 J/goC
T = to be calculated

Let the final temperature be T oC
use:
heat lost by 2 = heat gained by 1
m2*C2*(T2-T) = m1*C1*(T-T1)
37.8*0.449*(51.3-T) = 50.0*4.184*(T-25.0)
16.9722*(51.3-T) = 209.2*(T-25.0)
870.6739 - 16.9722*T = 209.2*T - 5230
T= 27.0 oC

Answer: 27.0 oC