Answer to Question #183311 in General Chemistry for jjjj

Question

When 1.00 kg lead (specific heat = 0.13 J / g °C)at 100 °C is added to a quantity of water at 28.5 °C, the final temperature of the lead – water mixture is 35.2 °C. What is the mass of water present?

Answer 1

Heat lost by lead is equal to the heat gained by water (assuming no heat loss to the surroundings). The heat lost by lead can be found directly:

$Q_{lead}=c_{lead}m_{lead}\Delta{T_{lead}}=0.13J/g\degree{C}\times1000g\times(100\degree{C}-35.2\degree{C})=8424J$

The specific heat of water is known to be 4.184 J/g^oC.

Therefore,

$m_{water}=\frac{Q_{water}}{c_{water}\Delta{T_{water}}}=\frac{8424J}{4.184J/g\degree{C}\times(35.2\degree{C}-28.5\degree{C})}=301g$

Answer: 301 g