Themo-Chemistry Related

Question

A 20.0 g block of iron (molar heat capacity 25.1 J/mol･°C) at 50.0 °C is placed into 200.0 g of water initially at 25.0 °C. What is the change in temperature (in °C) of the iron block? (The molar heat capacity of water is 75.38 J/mol･°C).

Answer 1

Heat LOST by the warm iron block MUST EQUAL heat gained by the cooler water.

heat lost by iron = q = mC∆T where m = mass; C = specific heat; ∆T = change in temperature

Since mass is in grams and heat capacity is in J/mol/degree, we will change units of heat capacity

25.1 J / moleº x 1 mol Fe / 55.85 g = 0.449 J/gº

heat lost by iron = (20.0 g)(0.449 J/gº)(50º - Tf) where Tf is the final temperature of the iron and water

heat gained by water = q = mC∆T

Since mass is in grams and specific heat is J/molº, we will change units of heat capacity

75.38 J/molº x 1 mol / 18 g = 4.18 J/gº

heat gained by water = (200.0 g)(4.18 J/gº)(Tf - 25.0º)

Setting heat lost by iron to heat gained by water and solving for final temp, we have...

(20.0 g)(0.449 J/gº)(50º - Tf) = (200.0 g)(4.18 J/gº)(Tf - 25.0º)

449 - 8.98Tf = 836Tf - 20,900

844.98Tf = 21,349

Tf = 25.27 This is the final temperature of the water and the iron block (equilibrium temperature)

The change in temperature of the iron block is thus

25.27º - 50.0º = -24.27º = -24.3º change in temperature of the iron block