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