all questionsanswerarrow-double-downarrow-downarrow leftarrow rightarrow upask parent to payattachmentboldbulleted listcameracheckclosecountercredit cardequationexcellentexclamation markfacebookfriendsheadingheartimageinstagramitaliclesslinkedinlock with playlogoutmediummenumessagesmicmoney transfermorenotificationsnumbered listpadlockpencilplaypluspointsprofileprofile_viewquestion markrecent questionsreloadreport flagrotatesearchseensettingssharesmsstarstar halfsubtitlesymbolstitletoughest questionstwitterunderlinedverifiedyoutube
Brainly.comBrainly.com
Join now
Previous question
Next question
brittnum3358 Asked 11.20.2017

A 16.2 g piece of aluminum (which has a molar heat capacity of 24.03 j/°c·mol) is heated to 82.4°c and dropped into a calorimeter containing water (specific heat capacity of water is 4.18 j/g°c) initially at 22.3°c. the final temperature of the water is 25.7°c. ignoring significant figures, calculate the mass of water in the calorimeter.

See answers (2)
  • Ask for details
  • Follow
  • Report

Answer Expert Verified
(get more)

5.0/5
0
shinmin
shinmin
florianmanteyw
florianmanteyw learned from this answer

The solution is:

q = m*c*dT 

First I converted 24.03J/°Cmol to grams 
(24.03J/°Cmol) * (1 mol Al/27g Al)= 0.89J/°Cg 

The final temperarure of water will be similar as the final temperature of aluminum 

q(aluminum) = 16.2g * 0.89J/°Cg * (25.7C - 82.4°C) = -817.5006 J 

q(water) = +817.5006 J = m * 4.18J/g°C * (25.7°C - 22.3°C) 

817.5006 = 14.212 m 
m = 57.52 grams of water or almost 58 grams

 

2
answers left

Have more answers for free

Get 20 Expert-Verified answers daily
Learn with a massive community
Ask your own questions
+98
diavinad8 and 98 others just joined Brainly

Click to let others know, how helpful is it
5.0
1 vote
1 vote
Rate!
Rate!
  • Comments
  • Report

Answer

0
adityaso

The mass of water in the calorimeter is \boxed{57.57\text{ g}}.

Further Explanation:

Whenever two bodies come in contact they exchanges heat energy until the two finally attains thermal equilibrium (means attaining a similar final temperature).

Concept:

Whean two bodies at different temperature are kept in contact, heat starts to liberated from the body at higher temperature to the body at lower temperature. Finally the two bodies reach the stage of same temperature and they stops liberating heat and remain stable.

Now, here we have a container containing water of some mass at 22.3^\circ\text{ C} to which we add a piece of aluminum of mass 16.2\text{ g}. Now aluminum, being at higher temperature will lose heat to the water.

The heat lost by the piece of aluminium to the water is given by:

\Delta{Q_{Al}={m_{Al}}{C_{S(Al)}\Delta T                                       …………. (1)

Here, m_{Al} is mass of aluminum, Q_{Al} is the heat liberated by aluminium.

The temperature difference from initial to final  temperature is:

\Delta T={T_\text{final}}-{T_\text{initial}}

Here, C_{S(Al)} is specific heat capacity of aluminum.

So, equation (1) can also be written as.

\Delta {Q_{Al}}={m_{Al}}{C_{S({Al})}}({{T_\text{final}}-{T_\text{initial})                                       …… (2)

Now, from equation (1), we have:

{m_{Al}}{C_{S\left({Al}\right)}}=\dfrac{{\Delta {Q_{Al}}}}{{\Delta T}}                                     …… (3)

The molar heat capacity of Aluminum is given by  

n{C_{m(Al)}} = \dfrac{{\Delta {Q_{Al}}}}{{\Delta T}}                                     …… (4)

Now from equation (3) and (4), we have

{m_{Al}}{C_{S({Al})}}=n{C_{m(Al)}}

Here, n is no. of moles of aluminum.

The specific heat capacity of Aluminum is given as:

C_{S(Al)}=\dfrac{C_{m(Al)}}{M_{(Al)}}

Here, M_{Al} is the molar mass (mass of 1 mole of the substance) of aluminum.

\begin{aligned}{M_{Al}}{C_{S\left( {Al} \right)}}&={C_{m\left( {Al}\right)}}\\{C_{S\left( {Al} \right)}}&=\frac{{{C_{m\left( {Al}\right)}}}}{{{M_{Al}}}}\\\end{aligned}

Substitute the value of variables in above expression.

\begin{aligned}{C_{S\left({Al}\right)}}&=\dfrac{{24.03\,{\text{J/^\circ C}}\cdot{\text{mol}}}}{{26.98{\text{g/mol}}}}\\&=0.89\text{ J/}^\circ\text{C}\end{aligned}

Substitute the value of of the variables in equation (2).

\begin{aligned}\Delta{Q_{Al}}&=(16.2\text{ g})(0.89\text{ J/}^\circ{\text{C}\cdot\text{g})(25.7^\circ\text{ C}-82.4^\circ\text{ C})\\&=-817.50\text{ J}\end{aligned}

So, the heat lost from the aluminum is:

\Delta{Q_\text{lost}}=-817.50\text{ J}

Therefore, this is the net amount of heat lost by the aluminum when it’s been dipped in the container of water.

Now, water is absorbing the heat given by aluminum. So the heat absorbed is given by:

\begin{gathered}{Q_{water}} = {m_{water}}{C_{S\left( {water} \right)}}\Delta T \hfill \\{Q_{water}} = {m_{water}}{C_{S\left( {water} \right)}}\left( {{T_{final}} - {T_{initial}}} \right) \hfill \\ \end{gathered}

Substitute values of variables in above expression.

\begin{aligned}Q_{\text{water}}&=m_\text{water}(4.18{\text{ J/}^\circ\text{C}\cdot\text{g})(25.7^\circ\text{ C}-22.3^\circ\text{ C})\\&=m_\text{water}(14.2)\end{aligned}

Therefore, the heat gain by the water present in the container is given by:

{Q_{gain}} = 14.2\left( {{m_{water}}} \right)                                     …………. (6)

Now according to the principle of Calorimetry,  the total sum of heat loss from a substance and heat gain to another substance is always equal to zero; provided both substances are present in isolated substance .i.e. no heat exchange with the surroundings.

\boxed{\text{Heat loss}=\text{Heat gain}}

The total sum of heat loss from a substance and heat gain to another substance is always equal to zero; provided both substances do not exchange heat with the surroundings.

Therefore, the above expression can be written as,

{Q_{gain}} + {Q_{loss}} = 0                …………(7)

Now, from equations (5) and (6), substituting the values of {Q_{loss}} and {Q_{gain}} respectively,

{Q_{gain}}+{Q_{loss}}=14.2({{m_{water})+(-817.50{\text{ J})

From equation (7),

\begin{aligned}14.2\left({{m_{water}}}\right)+\left({-817.50{\text{J}}}\right)&=0\\14.2({m_{water}})&=817.5\\{m_{water}}&=\left({\frac{{817.5}}{{14.2}}}\right)\\&=57.57{\text{g}}\\\end{aligned}

 

Thus, the mass of water present in the calorimeter is \boxed{57.57\text{ g}}.

Learn more:

1.  To what volume will a 2.33 L sample of gas expand brainly.com/question/9979757

2.  Average translational kinetic energy brainly.com/question/9078768

Answer Details:

Grade: High school

Subject: Physics

Chapter: Heat and Thermodynamics

Keywords:

16.2g, piece of aluminum, piece of aluminium, 82.4 C, calorimeter, initial temperature, 22.3 C, 25.7 C, heated, 24.03J/C mol, mass of water, calorimetry, substance.

Click to let others know, how helpful is it
0.0
0 votes
0 votes
Rate!
Rate!
  • Comments
  • Report

Looking for something else?

Looking for something else?
Find more answers

You might be interested in

anthonydjfjf2096
Asked 07/31/2019
See answers (1)
coreycullen8698
Asked 07/18/2019
See answers (1)
RyderStephens
Asked 07/27/2017
See answers (2)
4.2/5
17
caliesofiams
Asked 11/26/2016
See answers (1)

New questions in Physics

A water reservoir tank of capacity 250cubic metre is situated at height of 20m from the water level. What will be the power of an electric motor to be
used to fill tank in the 3 hours? Effeciency of motor is 70%.
Note that the simulation allows you to also display the force of the smaller moon pulling upon the larger planet. Click/tap the button to show the for
ce acting upon the planet. Which force is largest: the force ON the smaller moon or the force ON the larger planet?
A mechanic uses a hydraulic car jack to lift the front end of a car to change the oil. The jack used exerts 8,915 N of force from the larger piston. T
o pump the jack, 444 N of force is exerted on the small piston, which has an area of 3.14 cm2. What is the area of the large piston?
An automobile traveling on a straight, level road has an initial speed v when the brakes are applied. In coming to rest with a constant acceleration,
it travels a distance x. How far would the automobile travel in coming to rest if it had the same acceleration but an initial speed 2v ?
Why does the frequency of a wave increase as the wavelength decreases? A) because the wave speed remains constant B) because the amplitude of the wave
must remain constant C) because the frequency must get larger to increase the speed D) because the wavelength must get smaller to decrease the speed Eliminate
If you fill up a balloon with a small amount of air, then set it in direct sunlight, you will see that the balloon expands. This expansion is due to a
n increase in volume, which results from an increase in the temperature of the air in the balloon. This increase in temperature increases the ______________ of the gas, which results in the gas particles moving faster and colliding more often with the wall of the container. These collisions are what increase the pressure of the gas on the balloon, and ultimately cause the balloon to stretch out. A) moles B) spin energy C) kinetic energy D) potential energy
I will GIVE BRAINLYEST AND POINTS PLEAS HELP!!! uses cultural movement activities may use camps to teach activities uses seasons and teams to educate
teaches concepts and self-management skills uses challenging recreational activities options 1.sport education 2.adventure education 3.outdoor education 4.dance education 5.fitness education
A dock worker loading crates on a ship finds that a 24 kg crate, initially at rest on a horizontal surface, requires a 65 N horizontal force to set it
in motion. However, after the crate is in motion, a horizontal force of 50 N is required to keep it moving with a constant speed. The acceleration of gravity is 9.8 m/s 2 . Find the coefficient of static friction between crate and floor.
Consider the human body as a system and apply the first law of thermodynamics to it. We know that over any given period of sufficient length (say one
day), there will be a net heat flow from the body (i.e. Q is negative) and the body will do some external work on its surroundings (i.e. W is negative). The first law then tells us that ΔU is negative. So, each day there would be a decrease in internal energy. a. Does the internal energy of the human body decrease as described above?
The atomic radius of main-group elements generally increases down a group because ________. A) effective nuclear charge decreases down a group B) effe
ctive nuclear charge zigzags down a groupC) the principal quantum number of the valence orbitals increases D) effective nuclear charge increases down E) a group both effective nuclear charge increases down F) a group and the principal quantum number of the valence orbitals increases