Solution
PrepMate
To determine the mass of water in a specific volume, we need to use the density of water and the volume in question. The density of water at room temperature (approximately 20 degrees Celsius or 68 degrees Fahrenheit) is typically 1 gram per cubic centimeter (g/cm³) or 1000 kilograms per cubic meter (kg/m³).
To determine the mass of water in a specific volume, we need to use the density of water and the volume in question. The density of water at room temperature (approximately 20 degrees Celsius or 68 degrees Fahrenheit) is typically 1 gram per cubic centimeter (g/cm³) or 1000 kilograms per cubic meter (kg/m³).
Let's assume we want to find the mass of water in a volume of 2 liters. Here are the steps to calculate the mass:
Step 1: Convert the volume from liters to cubic centimeters (cm³) or cubic meters (m³), depending on the units you want to use for the mass. Since there are 1000 cubic centimeters in a liter, we have:
For our example:
Step 2: Use the density of water to find the mass. Since the density of water is 1 g/cm³, the mass in grams is numerically equal to the volume in cubic centimeters. Therefore:
For our example:
Step 3: If you need the mass in kilograms, convert grams to kilograms by dividing by 1000, since there are 1000 grams in a kilogram:
For our example:
Therefore, the mass of water in a volume of 2 liters is 2000 grams or 2 kilograms.
Let's assume we want to find the mass of water in a volume of 2 liters. Here are the steps to calculate the mass:
Step 1: Convert the volume from liters to cubic centimeters (cm³) or cubic meters (m³), depending on the units you want to use for the mass. Since there are 1000 cubic centimeters in a liter, we have:
For our example:
Step 2: Use the density of water to find the mass. Since the density of water is 1 g/cm³, the mass in grams is numerically equal to the volume in cubic centimeters. Therefore:
For our example:
Step 3: If you need the mass in kilograms, convert grams to kilograms by dividing by 1000, since there are 1000 grams in a kilogram:
For our example:
Therefore, the mass of water in a volume of 2 liters is 2000 grams or 2 kilograms.
Exercise 3 - Calorimetric calculations of final temperature
Exercise 1 - Calculating the amount of heat required to change temperature
Exercise 17 - Calculating volume of solution (Density, concentration & mass are given)
This video is available only for
registered users.s
Unlock this answer now, try 7 day free trial.
Exercise 3 - Calorimetric calculations of final temperature
Solved: Unlock Proprep's Full Access
Gain Full Access to Proprep for In-Depth Answers and Video Explanations
Signup to watch the full video 05:53 minutes
Hi, we're going to solve the following exercise.
A 850 gram piece of copper at 58 degrees Celsius is added to
500 milliliters of water maintained at 35 degrees Celsius in an insulated container.
What will be the final temperature of the Cu-H_2O mixture?
The specific heat capacity of water equals 4.18 Joule per gram degrees Celsius.
The specific heat capacity of copper equals 0.385 Joules per gram degrees Celsius.
We can see from the question that the mass of the copper,
just going to write that here, equals 850 grams.
We also know that the initial temperature of the copper equals 58 degrees Celsius,
so Ti equals 58 degrees Celsius.
Now we know that the copper is added to water.
It's added to 500 milliliters of water.
The volume of the water equals 500 milliliters.
We also knew that the initial temperature of the water is 35 degrees Celsius.
T initial of water equals 35 degrees Celsius and before it was a T initial of copper.
Now we're looking for the final temperature of the copper water mixture.
In order to solve this problem,
we're going to use the equation q.
The quantity of heat equals ms Delta T,
m is the mass of the substance,
s is the specific heat of the substance.
Now, the symbol of the specific heat is usually given a lowercase c. In our case,
we're using an s just to avoid confusion with heat capacity,
which is an uppercase C,
we're going to see that in later problems,
times Delta T,
Delta T equals T final minus T initial.
Now the quantity of heat which is lost by the copper,
is absorbed by the water.
Therefore, the quantity of heat of the copper plus
the quantity of heat of the water equals 0.
Again, q equals ms Delta T. The mass of the copper,
times the specific heat of the copper,
times Delta T of the copper,
plus the mass of the water times the specific heat of
the water, times Delta T of the water equals 0.
The mass of the copper, we said,
is 850 grams, times the specific heat of copper,
which is 0.385 Joule
per gram degrees Celsius times Delta T. Delta T is T final,
which we don't know.
That's what we want to find out, minus T initial,
which equals 58 degrees Celsius.
That's the quantity of heat,
the q of the copper.
Now we're adding the quantity of heat of the water.
The quantity of heat of the water is the mass of the water.
Now, remember the water is given in volume.
The volume of the water equals 500 milliliters.
We're just quickly going to calculate the mass.
Remember that d, the density,
equals m divided by v,
which is the mass divided by the volume.
Here we're looking for the mass, we're going to
multiply both sides by the volume, so m,
the mass equals d times v, density times volume.
In our case, the mass of the water equals the density of water,
which is 1 gram per milliliter,
times the volume of the water,
which is 500 milliliters,
so times 500 milliliters.
The milliliters cancel out and we're left with 500 grams.
The mass of the water is 500 grams.
This is times the specific heat of the water,
which is 4.18 Joule per gram degrees Celsius
times Delta T. Again,
Delta T of the water is T final, which we don't know.
That's what we're trying to find out,
minus the initial temperature of the water,
which is 35 degrees Celsius,
so minus 35 degrees Celsius.
All of this equals 0.
We can see that the grams cancel out here.
We're multiplying in here,
they also cancel out.
Now we will begin with multiplication.
This comes out to 327.25 Joule divided by degrees Celsius because that's
what we have leftover times
the T final minus 18,980.5.
In here, in our second component,
we're only left with Joules because the degrees Celsius and degrees Celsius cancel out.
Plus 2090 Joule per degrees Celsius times Tf minus
73,150 Joule.
All of this equals 0. We're going,
to sum up, the components of the T final,
and this comes to 2417.25
Joule divided by degrees Celsius times Tf.
On the other side, we have the Joule components,
which comes out to 92,130 Joule.
The Joules cancel out,
and Tf comes out to 38.11 degrees Celsius.
This is approximately 38 degrees Celsius.
The final temperature that we calculated equals 38 degrees Celsius.
That is our final answer.
Thank you very much for watching.
This video explains how to solve an exercise involving the addition of a 850 gram piece of copper at 58 degrees Celsius to 500 milliliters of water maintained at 35 degrees Celsius in an insulated container. The final temperature of the Cu-H_2O mixture is calculated using the equation q = msΔT, where m is the mass of the substance, s is the specific heat of the substance, and ΔT is the change in temperature. The final temperature of the mixture is 38.11 degrees Celsius.
Ask a tutor
If you have any additional questions, you can ask one of our experts.
Recently Asked Questions
