Question

calorimetry

Answer 1

125ml of water origianlly at 75.00 degrees are mixed with 225ml of water, originally at 25.00 degrees. Calculate final temperature of this mixture.

Answer 2

im not sure what to do or how to go about to do this....

Answer 3

like is one water the substance and one the surrounding? or are they both the surrounding? idk

Answer 4

According to the law of thermodynamics, when two substances get in contact, the final temperature is half of the sum of them.i in this case the final temperature should be (75+25)/2=50degreesC.

Answer 5

of the same substances coming in contact? what if it was like copper immersed in water?

Answer 6

ok but back to the old question, is there any other way you could show me? since this is calorimetry, what about the formulas and what not like q=mc(delta)t?

Answer 7

@Hoslos is incorrect. Use the principle of heat exchange:
\[\color{green}{-E_1 = E_2 \\ -m_1c\Delta T_1 = m_2c\Delta T_2 \\ -(125)c (T_f - 75.00) = (225)c(T_f - 25.00)\\-125T_f +(125)(75.00) = 225T_f - (225)(25.00)\\ T_f = \dfrac{1.50 \times 10^4}{350}\\T_f = 42.9~^{\circ}\text{C}}\]

Answer 8

In my solution above, note that the specific heat capacity, c, cancels out. Also, I have used a density of water of 1.00 g/mL. The density of water varies slightly with temperature, but this does not change the answer significantly, making the simplification possible.

Answer 9

I think you got that right.

Answer 10

ok thanks @ghuczek but why is @hoslos wrong then?

Answer 11

and next question

A 100g copper spoon is immersed in 500ml of water, originally at 90.00 degrees celcius. If the final temp of system reaches 88.55 dgs Celcius, what is the original temp of spoon?

So what i know so far is that water is going to be transferring the thermal energy to the spoon, therefore an increase in temp of spoon.

So i wrote down,...\[-(500g)(4.19)(tf-90.00 dgs celcius) = (100g)(0.385)(88.55-ti)\]

Answer 12

but how do you get tf?

Answer 13

You don't have this set up correctly. For the water\[\color{green}{\Delta T = T_f - T_i \\ = 88.55~^\circ\text{C} - 90.00~^\circ\text{C}}\] Then solve for the initial temperature of the spoon algebraically. That's what you are trying to find. You know the initial temperature of the water and the final temperature of the system.

Answer 14

But why are you using the temp of both?

Answer 15

The thermal energy lost by one object is equal to the thermal energy gained by the other object in an isolated system.

Answer 16

i still dont get it, the change in temp was -1.45 but now what? is the change in temp for both water and spoon?

Answer 17

this is what i did \[(100g)(0.385)(88.55-ti)= -(500g)(4.19)(-1.45)\]

gt like 9.64 or something so i know for a fact that im wrong

Answer 18

The final temperature will end up between the initial temperatures of the two things bing mixed together. 9.64, or something like that is not reasonable.