Thermochemistry

Question

A ball bearing at 259.0 °C is dropped into a cup containing 246.1 g of water at 17.7 °C. The water and ball bearing come to a temperature of 36.3 °C. What is the heat capacity of the ball bearing in J/°C?

C =    J/°C

Answer 1

ΔT = change in temperature, ºC

c = specific heat, J/(gºC)

q = heat, J

m = mass

q = mcΔT

heat capacity = q/ΔT = C

ΔT (water) = Tf - Ti = 36.3ºC - 17.7ºC = 18.6ºC

ΔT (bearing) = Tf - Ti = 36.3ºC - 259.0ºC = -222.7ºC

q (water) + q (bearing) = 0

mcΔT (water) + mcΔT (bearing) = 0

(246.1g)(4.184J/g°C)(18.6°C) + mc(-222.7°C) = 0

19152.09 J + mc(-222.7°C) = 0

mc(-222.7°C) = -19152.09 J

mc = C = 86.0 J/°C = heat capacity of the ball bearing

Answer 2

If we assume (which we have to) that no heat is lost to the surroundings, all of the energy lost by the ball bearing is absorbed by the water. So, using q = cmΔT, we solve for the q H2O. (Incidentally, they should have provided you with c of the water, which is 4.184J/g.^oC.) Then, we apply the same equation for the ball bearing. However, they didn't provide the mass of the ball bearing, so they want us to solve for heat capacity (extensive property), rather than the specific heat capacity (intensive property). C = cm = q/ΔT.

Incidentally, I've assumed you would use the absolute value of the temperature change. What's really going on is that the q lost by the ball bearing is equal to in magnitude and opposite in sign to the q gained by the water. 0 = q_H2O + q_bb And so, q_H2O = -q_bb

Answer 3

Heat capacity is the amount of heat needed to raise the temperature of an object by 1 degree.

We can find the heat involved in this problem by looking at the water with a mass of 246.1 g, a specific heat of 4.184 J/g/deg and a change in temperature of 18.6 degrees (going from 17.7º to 36.3º)

Heat = mC∆T = (246.1 g)(4.184 J/g/deg)(18.6 deg) = 19,152 J

Heat capacity of ball bearing = heat/∆T

heat = 19,152 J

∆T = 259.0 - 36.3 deg = 222.7 degrees

Heat capacity = 19,152 J / 222.7 degrees = 85.999 J/ºC = 86.00 J/ºC