Question

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.

Answer 1

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

 

Answer 2

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:

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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.