Question

7. A coffee cup calorimeter contains 150 g of water at 25.1 °C. A 121 g block of solid copper is placed into the water and the temperature of the water and block goes to 30.1 °C. What was the perature of the copper block? The specific heat of solid copper metal is 0.385 J/g-°C and the specific heat of water is 4.184 J/g-°C. Assume that the calorimeter doesn't absorb any heat.

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 = 121 g

C = specific heat of Cu = 0.385 J/g•C

T2 = final temp = 30.1°C

T1 = temperature of Cu metal = ?

Thus,

Qm = -121 g x 0.385 J/g°C x ( 30.1 °C - T1 ) ----------------(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 = 150 g

T1 = 25.1 °C

T2 = 30.1 °C

Hence,

Qw = 150 g x 4.184 J /g °C x ( 30.1 °C - 25.1 °C )  

Qw = 3765.6 J-------------------------------(2)

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

Hence, from equation 1 and 2,

-121 g x 0.385 J/g°C x ( 30.1 °C - T1 ) = 3765.6 J

(30.1 °C - T1)  = - 3765.6 / 46.585

(30.1 °C - T1) = - 80.8 °C

T1 = 110.9 °C