A150 g copper bowl contains 210 g of water, both at 24.0°c. a very hot 430 g copper cylinder is dropped into the water, causing the water to boil, with 17.1 g being converted to steam. the final temperature of the system is 100°c. neglect energy transfers with the environment.
(a) how much energy is transferred to the water as heat?
(b) how much to the bowl?
(c) what is the original temperature of the cylinder?
the specific heat of water is 1 cal/g·k, and of copper is 0.0923 cal/g·k. the latent heat of vaporization of water is 539 cal/kg.

A.15969.22 cal

B.1052,22 cal

C.528,87 °C

Explanation:

To solve this kind of question, a proper method is to work from the data that you have towards the data that you need. Also, it is recommended to analyze related equations as they could give us clues on how to find the missing information or the information that the problem is asking us.

Let us start with Question A. It is important to remember that energy transfers with the environment are being neglected; this means that all the energy that the cylinder lose is picked up by the water and the copper bowl. To find the amount of energy transferred to the water, we first find the amount of energy necessary to raise the water’s temperature to 100°C and then we find the amount of energy necessary to evaporate the 17.1 g of water indicated by the question. This would be:

Q = m_water * CP_water *∆T =210g *1 cal/(g K) * (100°C-24°C) = 15960 cal

Q_evap = m_wat * L = 17,1 g * 539 cal/kg* (1 kg)/(1000 g) =9.2169 cal

Therefore, the total energy that was transferred to the water is the sum of these components, that would be Q_tot = 15960 cal + 9.2159 cal = 15969.22 cal.  Let´s also remember that a temperature difference in K is equal to a temperature difference in ° C

To solve Question B, we use the same method. We must find the amount of energy necessary to raise the temperature from its initial temperature to the one stated by the problem to be the equilibrium temperature of the system (100°C):

Q= m_copper *CP_copper *∆T = 150g * 0.0923 cal/(g K) * (100°C-24°C) = 1052,22 cal

If we add the components we just found in questions A and B, we can find the amount of energy than the Copper cylinder lost, this would be: Q_tot = 15969.22 cal + 1052.22 cal = 17021.44 cal.

The question C asks us to find the initial temperature of the cylinder and Q_tot will help us to find it.

We know that Q_tot is the energy lost by the cylinder and we also know that Q_tot = m_cylinder * CP_copper * ∆T. Therefore, what we need to do  is clear the last term of the equation and find the initial temperature.

Q_tot = m_cylinder *CP_copper *∆T → T_fin-T_initial = Q_tot/(m_cylinder*CP_copper ) = (-17021.44 cal)/(430g*0.0923 cal/(g K))

→ T_initial = 100°C + (-17021.44 cal)/(430g * 0.0923 cal/(g K)) = 528,87 °C

If we convert the 100°C to K before we do the calculation, the result would be the same one, You would only need to add 273,15 to the final result to check it out.  

Hope everything was clear. If you have any further question, I'll be happy to help :D

the bal will not change its velocity.