answer the following

Question

A 30.14-g stainless steel ball bearing at 117.82°C is placed in a constant-pressure calorimeter containing 120.0 mL of water at 18.44°C. If the specific heat of the ball bearing is 0.474 J/gC°, calculate the final temperature of the water. Assume the calorimeter to have negligible heat capacity

Answer 1

Heat LOST by steel MUST equal heat GAINED by water.

Heat lost by steel = q = mC∆T

q = heat = ?

m = mass = 30.14 g

C = specific heat = 0.474 J/gº

∆T = change in temperature = 117.82º - Tf (Tf = final temperature)

q = (30.14 g)(0.474 J/gº)(117.82º - Tf) = 1683 - 14.29 Tf

Heat gained by water = q = mC∆T

q = heat = ?

m = mass = 120.0 g (assuming a density of 1 g / ml)

C = specific heat = 4.184 J/gº

∆T = change in temperature = Tf - 18.44º)

q = (120.0 g)(4.184 J/gº)(Tf - 18.44º) = 502Tf - 9258

Set heat lost by steel equal to heat gained by water and solve for Tf:

1683 - 14.29Tf = 502Tf - 9258

10941 = 516Tf

Tf = 21.1º = FINAL TEMPERATURE

(be sure to check all of the math)