Question

A 39.9 gram piece of lead at 98.7 °C is placed in a calorimeter containing water at 21.7 °C. If the temperature at equilibrium is 27.6°C, what is the mass of the water?

Answer #1

*Given data,*

*mmetal = 39.9 g,*

*Tf = 27.6°C ,*

*Tmetal = 98.7 °C*

*We know , C _{lead} = 0.13 J/gC, C_{water} =
4.184 J/gC*

*Let us consider,*

*-Qlost = Qgain , -Qmetal = Qwater*

*Qmetal = mmetal * Cmetal * (Tf-Tmetal)*

*Qwater = mwater * Cwater * (Tf - Twater)*

*- mmetal * Cmetal * (Tf-Tmetal) = mwater * Cwater * (Tf -
Twater)*

*Substitute these values in above equation,*

*-39.9 * 0.13(27.6°C - 98.7 °C) = mwater * 4.184*(27.6 -
21.7)*

*368.79 = mwater * 5.9*

**m _{water} = 62.5 g**

*Therefore, the mass of water = 62.5 g*

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