Question

Part A Two substances, lead and sand, initially at different temperatures, come into contact and reach thermal equilibrium. The mass of lead is 19.2 g and its initial temperature is 57.9 ∘C. The mass of sand is 28.8 g and its initial temperature is 27.6 ∘C. What is the final temperature of both substances at thermal equilibrium? (The specific heat capacity of lead is 0.128 J/g⋅∘C; the specific heat capacity of sand is 0.84 J/g⋅∘C.) Express your answer to three significant figures. A 6.78 g sample of gold, initially at 74.7 ∘C, is submerged into 88.7 g of ethanol at 12.4 ∘C in an insulated container. What is the final temperature of both substances at thermal equilibrium? (The specific heat capacity of gold is 0.128 J/g⋅∘C; the specific heat capacity of ethanol is 2.42 J/g⋅∘C.) Express your answer to three significant figures.

Answer 1

heat lose of lead                            =               heat gain of sand

mcT                                           =                mcT

19.2*0.128*(57.9-t)                        =                28.8*0.84*(t-27.6)

           t            = 30.4⁰C

The final temperature = 30.4⁰C

part-B

heat lose of gold                     =                 heat gain of ethanol

mcT                                    =                mcT

6.78*0.128*(74.7-t)                 =                88.7*2.42*(t-12.4)

             t   = 12.7⁰C

The final temperature is 12.7⁰C