Question

1) A 32 g stainless steel ball at 110.5°C is placed in a constant pressure calorimeter containing 100 ml of water at 20 °C. Calculate the final temperature of water. The specific heat of the ball is 0.474 J/g°C and the specific heat of water is 4.18 J/g°C.

explain plz

Answer 1

Let us denote water by symbol 1 and ball by symbol 2

since volume of water is 100 mL and density is 1 g/mL,mass of water = 100 g

m1 = 100 g

T1 = 20 oC

C1 = 4.18 J/goC

m2 = 32 g

T2 = 110.5 oC

C2 = 0.474 J/goC

Let the final temperature be T oC

we have below equation to be used:

heat lost by 2 = heat gained by 1

m2*C2*(T2-T) = m1*C1*(T-T1)

32.0*0.474*(110.5-T) = 100.0*4.18*(T-20.0)

15.168*(110.5-T) = 418*(T-20.0)

1676.064 - 15.168*T = 418*T - 8360

T= 23.2 oC

Answer: 23.2 oC