A 5.00g ball of hot iron at 200 degrees celcius is dropped int owater at 20 degrees celcius. The system acheives thermal equilibrium at 25 degrees celicus. calculate the mass of water heated.

the mass of water heated is m water = 18.81 g

Explanation:

Assuming a specific heat of iron of C iron = 0.450 J/g°C (from tables), then if we neglect heat losses to the environment, all the heat absorbed by the water is the one released by the iron ball.

Therefore

heat released = Q = m iron * C iron * (T initial iron - T final) = 5 g * 0.450 J/g°C (200°C - 25°C) = 393.75 J

where

m = mass, T = temperature

then

heat released = heat absorbed = Q = m water * C water * (T final - T initial water)

m water = Q / [C water * (T final - T initial water) ] = 393.75 J/[ 4.186 J/g°C * (25°C-20°C) ] = 18.81 g

m water = 18.81 g