How Much Energy Does Dropping a Hot Copper Cylinder into Water Transfer?

[sciontc03]

A 125 g copper bowl contains 235 g of water, both at 20.0°C. A very hot 300 g copper cylinder is dropped into the water, causing the water to boil, with 5.05 g being converted to steam. The final temperature of the system is 100°C. Neglect energy transfers with the environment.
How much energy is transferred to the water as heat?

I found the heat for the steam, bowl, water, and part of the cylinder except for the initial temp of the cylinder. I don't know where to go from there.
Qs= 11.39kJ
Qw=78.772kJ
Qb=3.86kJ
Please help.

 

[jbsteele]

The total heat transferred to the water is the sum of the heats of the steam, bowl and part of the cylinder. This is equal to 94.031 kJ.

 

[baba89]

Based on the given information, we can calculate the initial temperature of the copper cylinder by using the conservation of energy equation: Qs + Qw + Qb = Qc, where Qc is the heat transferred to the copper cylinder.

Qc = m x c x ΔT (mass x specific heat capacity x change in temperature)

Qc = (300g)(0.385 J/g°C)(Tf - Ti)

Qc = 115.5Tf - 115.5Ti

We know that the final temperature (Tf) is 100°C, and the initial temperature (Ti) is unknown. We also know that the heat for the steam (Qs) is 11.39 kJ, the heat for the water (Qw) is 78.772 kJ, and the heat for the bowl (Qb) is 3.86 kJ.

Substituting these values into the conservation of energy equation, we can solve for the initial temperature of the copper cylinder:

11.39 kJ + 78.772 kJ + 3.86 kJ + 115.5Tf - 115.5Ti = 0

115.5Tf - 115.5Ti = -94.022 kJ

Since we know the final temperature (Tf) is 100°C, we can solve for the initial temperature (Ti):

115.5(100) - 115.5Ti = -94.022 kJ

11550 - 115.5Ti = -94.022 kJ

115.5Ti = 11644.022 kJ

Ti = 100.7°C

Therefore, the initial temperature of the copper cylinder was 100.7°C before it was dropped into the water.

To calculate the total energy transferred to the water as heat, we can use the specific heat capacity of water (4.184 J/g°C) and the change in temperature (80°C):

Qw = m x c x ΔT

Qw = (235g)(4.184 J/g°C)(80°C)

Qw = 7817.04 J = 7.817 kJ

So, the total energy transferred to the water as heat is 7.817 kJ.