Final Temp of Water in Calorimeter After Heating Silver Ring

[laker88116]

A 25.5 gram silver ring ( c=234 J/kg * c) is heated to a temperature of 84 degrees C and then placed in a calorimeter containing 0.05 kg of water at 24 degrees C. The calorimeter is not perfectly insulated and .140 kJ of energy is transferred to the surrounding before a final temperature is reached. What is the final temperature?

Ok, I was thinking that there is no phase change so I am not sure is it just conservation of energy, and if so, how do I start?

 

[laker88116]

Nevermind, I was able to figure it out.

 

[sir-pinski]

Yes, you are correct that there is no phase change involved in this scenario. To find the final temperature, we can use the principle of conservation of energy. This principle states that the total energy in a system remains constant, it just changes form from one type to another.

In this case, we have two main sources of energy - the heat energy from the silver ring and the heat energy from the water in the calorimeter. The heat energy from the silver ring can be calculated using the formula Q = mcΔT, where Q is the heat energy, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.

So, the heat energy from the silver ring can be calculated as Q1 = (25.5 g)(234 J/kg * C)(84 C - Tf), where Tf is the final temperature of the system.

The heat energy from the water in the calorimeter can be calculated as Q2 = (0.05 kg)(4186 J/kg * C)(Tf - 24 C).

Since energy is conserved, we can equate Q1 and Q2 and solve for Tf:

Q1 = Q2

(25.5 g)(234 J/kg * C)(84 C - Tf) = (0.05 kg)(4186 J/kg * C)(Tf - 24 C)

After solving for Tf, we get a final temperature of 33.8 degrees Celsius. However, we also need to take into account the heat lost to the surroundings, which was given as 0.140 kJ. This heat energy will cause a slight decrease in the final temperature.

To calculate the final temperature taking into account the heat lost to the surroundings, we can use the formula Qlost = mcΔT, where Qlost is the heat energy lost, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.

So, the heat energy lost can be calculated as Qlost = (0.05 kg)(4186 J/kg * C)(33.8 C - Tf).

Equating this with the given value of 0.140 kJ, we can solve for Tf and get a final temperature of 33.9 degrees Celsius.

Therefore, the final temperature of the water in the calorimeter after heating the silver ring and accounting for heat loss to the surroundings is approximately 33.9 degrees Celsius.