Laboratory technician drops a 0.0850kg sample of unknown solid

  • Thread starter Thread starter Larrytsai
  • Start date Start date
  • Tags Tags
    Laboratory Solid
Click For Summary

Discussion Overview

The discussion revolves around a calorimetry problem involving the calculation of the specific heat of an unknown solid after it is dropped into a calorimeter containing water and copper. Participants explore the principles of heat transfer, specifically focusing on the heat gained by the water and the heat lost by the solid.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant states the principle of conservation of energy as Qsys = -Qsurr but expresses confusion regarding the specific heat variables involved.
  • Another participant suggests using the concept of heat gain equals heat lost, indicating that specific heats of copper and water should be referenced from tables.
  • A participant mentions the need to consider the specific heat of water in the context of the problem, implying familiarity with its properties.
  • In a separate post, a participant describes a similar calorimetry problem involving a stainless steel ball bearing and emphasizes the heat balance equation with one unknown temperature.
  • Another participant questions whether the change in temperature for water equals the change in temperature for the ball, seeking clarification on the relationship between heat transfer and temperature change.
  • A later reply asserts that the changes in temperature are directly proportional to the amount of heat transferred, but does not resolve the earlier question about equality of temperature changes.

Areas of Agreement / Disagreement

Participants generally agree on the principle of heat transfer but express differing views on the specific application of the equations and the relationships between temperature changes and heat transfer. The discussion remains unresolved regarding the specific heat calculation and the equality of temperature changes.

Contextual Notes

Participants reference specific heats of materials and the need for accurate values, indicating potential limitations in the assumptions made about the calorimeter's capacity and the specific heat values used.

Who May Find This Useful

This discussion may be useful for students or individuals studying calorimetry, heat transfer, and specific heat calculations in physics or chemistry contexts.

Larrytsai
Messages
222
Reaction score
0
A laboratory technician drops a 0.0850kg sample of unknown solid material, at a temperature of 100.0 degree celsius, into a calorimeter. The calorimeter can, initially at 19.0 degrees celsius, is made of 0.150kg of copper and contains 0.200kg of water. The final temperature of the calorimeter can and contents is 26.1 degrees celsius. Compute the specific heat of the sample.

The only thing i have so far is

Qsys= -Qsurr
but I am stumped with the 2 specific heat variables
 
Physics news on Phys.org


Not sys and surr, I would rather go for gain=lost.

I suppose you should check both specific heats of copper and water in tables.

--
 


Being calories and degrees C he ought to know that for water roughly.
 


Borek said:
Not sys and surr, I would rather go for gain=lost.

I suppose you should check both specific heats of copper and water in tables.

--
ChemBuddy chemical calculators - buffer calculator, stoichiometry calculator
www.ph-meter.info - ph meter, ph electrode


i'll be so grateful if you can help me for this problem:
a 30.14-g stainless steel ball bearing at 117.82 c is placed in a constant-pressure calorimeter containing 120.0 mL of water at 18.44 C.if the specific heat of the ball bearing is 0.474 J/g.c, calculate the final temperature of the water.assume the calorimeter to have negligible capaity.
 


It is again simple heat gain (by water) equals heat lost (by ball). Assume final temperature to be Tfinal and write equation for a heat balance. You will get equation with one unknown. That's all.

--
methods
 


Borek said:
It is again simple heat gain (by water) equals heat lost (by ball). Assume final temperature to be Tfinal and write equation for a heat balance. You will get equation with one unknown. That's all.

--
chemical calculators - buffer calculator, concentration calculator
www.titrations.info - all about titration methods


thank you so much,,i got the idea, but I'm still thinking about if q(water) equals q(ball), can we
prove that (Delta Temp) for water equals (Delata Temp) for the ball!
 


No, but both changes in temperature are directly proportional to the amount of heat transferred.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
Replies
1
Views
4K
Replies
2
Views
6K
Replies
3
Views
5K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 19 ·
Replies
19
Views
3K
Replies
6
Views
16K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
6K