Answer to Question #183311 in General Chemistry for jjjj

Question #183311

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?


1
Expert's answer
2021-04-23T07:30:22-0400

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/goC.

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


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