Thermodynamics: Calorimeter problem

In summary, to find the Ls (latent heat of steam), you need to set up an equation that equates the heat flow into the water and calorimeter to the heat flow out of the steam. This will involve calculating the heat lost as the steam condenses from 100°C to 50°C, in addition to the latent heat of condensation. Simply plugging in the latent heat of steam (2.26*10^6) is incorrect.
  • #1
petefic
3
0

Homework Statement


A 50.0g copper calorimeter contains 250g of water at 20 deg celsius. How much steam must be condensed into the water if the final temperature of the system is to reach 50 deg celsius?

Homework Equations


Can someone explain how to find the Ls? I was told that just plugging in the latent heat of steam (2.26*10^6) is incorrect.

The Attempt at a Solution



Qsteam = Qcalorimeter + Qwater
ms * Ls = (mc * cc * dT) + (mw * cw * dT)
ms * Ls = (0.05kg * 0.0924cal/g * (50deg-20deg)) + (.25kg * 1cal/g *(50deg-20deg))
ms * Ls = 7638.6cal
 
Last edited:
Physics news on Phys.org
  • #2
petefic said:

Homework Statement


A 50.0g copper calorimeter contains 250g of water at 20 deg celsius. How much steam must be condensed into the water if the final temperature of the system is to reach 50 deg celsius?

Homework Equations


Can someone explain how to find the Ls? I was told that just plugging in the latent heat of steam (2.26*10^6) is incorrect.

The Attempt at a Solution



Qsteam = Qcalorimeter + Qwater
ms * Ls = (mc * cc * dT) + (mw * cw * dT)
ms * Ls = (0.05kg * 0.0924cal/g * (50deg-20deg)) + (.25kg * 1cal/g *(50deg-20deg))
ms * Ls = 7638.6cal
You have to assume that the steam is at 100°C.

You have the right idea in setting up an equation that equates the heat flow into the 250 g of water and 50 g calorimeter to the heat flow out of an unknown mass of steam. But the heat flow from the steam occurs in two stages: 1. the release of the latent heat of condensation and 2. the heat lost in reducing its temperature from 100C to 50C.

AM
 

1. What is a calorimeter and how does it work?

A calorimeter is a device used to measure the amount of heat released or absorbed during a chemical or physical process. It typically consists of a container filled with a known amount of water, an insulated chamber to prevent heat from escaping, and a stirrer to ensure even temperature distribution. The heat released or absorbed by the substance being tested causes a change in the temperature of the water, which can be measured and used to calculate the heat exchanged.

2. What is the purpose of a calorimeter in thermodynamics?

In thermodynamics, a calorimeter is used to measure the heat of a reaction or the specific heat capacity of a substance. This information is important in understanding the energy changes that occur during a chemical reaction or physical process, and how these changes affect the overall system.

3. How do you set up a calorimeter for a calorimeter problem?

To set up a calorimeter for a calorimeter problem, you will need to measure the mass of the substance being tested, the initial temperature of the water in the calorimeter, and the final temperature after the reaction or process has occurred. You will then need to calculate the heat exchanged using the equation q = mcΔT, where q is the heat exchanged, m is the mass of the substance, c is the specific heat capacity of the substance, and ΔT is the change in temperature of the water. Make sure to insulate the calorimeter to prevent heat from escaping during the experiment.

4. How does a calorimeter problem relate to the first law of thermodynamics?

The first law of thermodynamics states that energy cannot be created or destroyed, only transferred or converted from one form to another. In a calorimeter problem, the heat released or absorbed during a reaction or process is a form of energy that is being transferred. By measuring this heat and using it to calculate the change in internal energy of the system, we can demonstrate the first law of thermodynamics.

5. What are some common sources of error in a calorimeter experiment?

There are several sources of error in a calorimeter experiment, including heat loss to the surroundings, incomplete combustion of the substance being tested, and heat losses due to friction from the stirrer. It is important to control for these sources of error by insulating the calorimeter and ensuring complete combustion of the substance, as well as accounting for any heat losses in the calculations.

Similar threads

  • Introductory Physics Homework Help
Replies
7
Views
12K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
7K
Back
Top