Chap 6 4. A 31.15g stainless steel ball bearing at 103.27°C is placed in a const

Question

Chap 6 4. A 31.15g stainless steel ball bearing at 103.27°C is placed in a constant-pressure calorimeter containing 116.9 g of water at 22.87°C. If the specific heat of the ball bearing is 0.474 J / (g · °C), calculate the final temperature of both the water and steel when they equilibrate. Assume the calorimeter to have negligible heat capacity. Chap 6 4. A 31.15g stainless steel ball bearing at 103.27°C is placed in a constant-pressure calorimeter containing 116.9 g of water at 22.87°C. If the specific heat of the ball bearing is 0.474 J / (g · °C), calculate the final temperature of both the water and steel when they equilibrate. Assume the calorimeter to have negligible heat capacity. 4. A 31.15g stainless steel ball bearing at 103.27°C is placed in a constant-pressure calorimeter containing 116.9 g of water at 22.87°C. If the specific heat of the ball bearing is 0.474 J / (g · °C), calculate the final temperature of both the water and steel when they equilibrate. Assume the calorimeter to have negligible heat capacity. 4. A 31.15g stainless steel ball bearing at 103.27°C is placed in a constant-pressure calorimeter containing 116.9 g of water at 22.87°C. If the specific heat of the ball bearing is 0.474 J / (g · °C), calculate the final temperature of both the water and steel when they equilibrate. Assume the calorimeter to have negligible heat capacity.

Explanation / Answer

Q = mcT

Q = heat energy (Joules, J), m = mass of a substance (kg)

c = specific heat (units J/kgK), is a symbol meaning "the change in"

T = change in temperature (Kelvins, K)

Heat lost by steel ball = heat gained by water

31.15 x 0.474 x 103.27°C - Tf = 116.9 x 4.184 x (Tf - 22.87)

1524.79 - 14.7651 Tf = 489.109 - 11185.936

503.8741 Tf = 13173.619

Tf = 26.14 Deg celcius

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