Grade 11 Heat Capacity question

[Confuzed_one]

Homework Statement

a 0.20kg lead shot is heated to 90 degrees celsius and dropped into an ideal calorimeter containing 0.50kg of water initially at 20 degrees celsius. What is the final equilibrium temperature of the lead shot? The specific heat capacity of lead is 128 Joules/(kg degrees celsius) and the specific heat capacity of water is 4186Joules/(kg degrees Celsius)

Ml = 0.20
Cl = 128
Til = 90

Mw = 0.50
Cw = 4186
Tiw = 20

Homework Equations

Q= mass x heat capacity x difference in temperature

Qabsorbed + Qreleased = 0

The Attempt at a Solution

0.20 x 128 x (T2 - 90) + 0.50 x 4186 x(T2 - 20)

25.6 x (t2 -90) + 2093 x (T2-20)

I am completely stuck after this. I don't know what to do with the difference in temperature part. I've looked at example equations and it shows the temperature given getting transformed into joules? So in this equation the 90 and 20 would be replaced by different numbers with the unit Joules after. I don't understand why this is happening. Thank you for any help and I appreciate you taking the time to read this.

This may not be very clear, so if I need to explain more please let me know

 

[gneill]

Homework Statement

a 0.20kg lead shot is heated to 90 degrees celsius and dropped into an ideal calorimeter containing 0.50kg of water initially at 20 degrees celsius. What is the final equilibrium temperature of the lead shot? The specific heat capacity of lead is 128 Joules/(kg degrees celsius) and the specific heat capacity of water is 4186Joules/(kg degrees Celsius)

Ml = 0.20
Cl = 128
Til = 90

Mw = 0.50
Cw = 4186
Tiw = 20

Homework Equations

Q= mass x heat capacity x difference in temperature

Qabsorbed + Qreleased = 0

The Attempt at a Solution

0.20 x 128 x (T2 - 90) + 0.50 x 4186 x(T2 - 20)

25.6 x (t2 -90) + 2093 x (T2-20)

I am completely stuck after this. I don't know what to do with the difference in temperature part. I've looked at example equations and it shows the temperature given getting transformed into joules? So in this equation the 90 and 20 would be replaced by different numbers with the unit Joules after. I don't understand why this is happening. Thank you for any help and I appreciate you taking the time to read this.

This may not be very clear, so if I need to explain more please let me know

Make an equation out of it as in your Relevant Equations section. Solve for T2.

When units (Joules, °C, etc.) get confusing, take your equation and substitute each item with its corresponding units. Check how the units cancel and combine. They should all 'harmonize' to produce the expected units that are on the other side of the equation.

 

[Confuzed_one]

I tried rearranging it to solve for T2, but I'm definitely doing something wrong.

0.20 x 128 x (t2 - 90)

-T2 = 0.20 x 128 - 90

-t2 = 25.6 - 90

-t2 = -64.4

T2 = 64.4

This is one of those multiple choice answer things, and this isn't an option sadly. I understand the whole unit thing, I'm just pressed for time and trying to shorten it. I really appreciate the help thank you

EDIT: I realized i should factor this, so I did it and I got the correct answer of 20.8 degrees celsius. I know this is kind of random and I am not explaining myself, but you really helped and I got the right answer. Thank you very much, you have no idea how much i appreciate it

 

[Delphi51]

May I offer a little clarification at the beginning?
Heat lost by lead = heat gained by water
mC*ΔT = MC*ΔT
0.2*128*(90-T2) = .5*4186*(T2-20)
This differs from your solution in that there is an equals sign in the middle
and your (T2-90) is replaced by (90-T2). (T2-90) would be a negative number, leading to all sorts of trouble!

Next step is to expand both sides out so you only T2 terms and numerical terms.
Collect like terms and solve for T2.
Keep the = sign in every line!

 

[asteriatic]

.Dear student,

Thank you for your question. It seems that you are on the right track with your attempt at solving the problem. The key concept to understand is that heat energy is transferred from the lead shot to the water in the calorimeter until they reach a final equilibrium temperature. This means that the amount of heat absorbed by the water must be equal to the amount of heat released by the lead shot. We can express this as an equation:

Qabsorbed = Qreleased

We can calculate the amount of heat absorbed by the water using the equation Q = m x c x ΔT, where m is the mass of the water, c is its specific heat capacity, and ΔT is the change in temperature. We can also calculate the amount of heat released by the lead shot using the same equation, but with the mass and specific heat capacity of lead and the difference in temperature between the initial temperature of the lead shot and the final equilibrium temperature.

Putting this all together, we get:

mwater x cwater x (Tf - Tiw) = mlead x clead x (Tf - Til)

Substituting the given values, we get:

0.50 x 4186 x (Tf - 20) = 0.20 x 128 x (Tf - 90)

Solving for Tf, we get:

Tf = (0.20 x 128 x 90 + 0.50 x 4186 x 20) / (0.20 x 128 + 0.50 x 4186) = 33.7 degrees Celsius

Therefore, the final equilibrium temperature of the lead shot is 33.7 degrees Celsius.

I hope this helps clarify the concept and guide you towards the solution. Keep up the good work and keep asking questions!

Best regards,