Question

A 43.0-g sample of copper at 99.9 °C is dropped into a beaker containing 149 g of water at 18.2 °C. What is the final temperature when thermal equilibrium is reached? (The specific heat capacities of liquid water and copper are 4.184 J/g · K and 0.385 J/g · K, respectively.) Final temperature =

Answer 1

Solution: The given problem can be solved by following relation,

Heat lost by Cu metal = Heat gain by water

Heat lost by metal (Qm) = - m C ΔT

Where, m = mass = 43 g

C = specific heat of Cu = 0.385 J/g•C

T2 = final temp = 18.2 °C

T1 = temperature of Cu metal = 99.9 °C

Thus,

Qm = - 43 g x 0.385 J/g°C x ( 18.2 °C - 99.9 °C )

Qm = 1352.5435 J  ----------------(1)

Heat gain by water (Qw),

Qw = m S ΔT

Where, S = specific heat of water = 4.184 J /g °C

m = mass of water = 149 g

T1 = 18.2 °C

T2 = ?

Hence,

Qw = 149 g x 4.184 J /g °C x ( T2 - 18.2 °C ) ------ (2)

Since, Heat lost by Cu metal (Qm) = Heat gain by water (Qw)

Hence, from equation 1 and 2,

149 g x 4.184 J /g °C x ( T2 - 18.2 °C ) = 1352.5435 J

(T2 - 18.2 °C)  = 1352.5435 J / 149 g x 4.184 J /g °C

(T2 - 18.2 °C) = 2.17 °C

T2 = 18.2 + 2.17 = 20.37 = 20.4 °C