5.6: Calorimetry
- Page ID
- 167656
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Ideal Calorimetry
Exercise \(\PageIndex{1}\)
If a 55.8 g piece of silver at 75.8°C is placed in 150.0 g of water at 35.2°C, what is the final temperature? Assume that no heat is transferred to the surroundings.
- Answer
-
36.0 °C
\((mc_{s}(T_{final}-T_{initial}))_{Ag} + (mc_{s}(T_{final}-T_{initial}))_{H_{2}O} = 0\)
\((55.8 g)(0.240 J/g*°C)(T_{final} - 75.8°C) + (150 g)(4.184 J/g*°C)(T_{final} - 35.2 °C) = 0\)
\(T_{final}(13.392 J/g*°C) -1015.1136 J + T_{final}(627.6 J/g*°C) - (22091.52 J) = 0\)\(T_{final}(640.992 J/g*°C) = 23106.6336 J\)
\(T_{final} = 36.048 °C\)
Exercise \(\PageIndex{2}\)
If a 105.3 g piece of copper pipe at 63.8°C is placed in 94.7 g of water at 27.0°C, what is the final temperature? Assume that no heat is transferred to the surroundings.
- Answer
-
30.4 °C
Real Calorimetry
Exercise \(\PageIndex{3}\)
If 229 g of tin at 110.0°C were placed in a calorimeter with 100 g of water at 21.0°C and the resulting temperature of the mixture was 26.4°C, calculate the qcal, qH20, qPb? Calculate the calorimeter constant (Cc) for the particular calorimeter that was used to collect the data. The specific heat of water is 4.184 J/g °C and the specific heat of tin is 0.213 J/g °C.
- Answer
-
qcal=1.82*103 J; qH2O = 2.259*103 J; qSn = -4.077*103 J; Cc = 336.7 J/°C
\(q = mc_{sp}\Delta T\)
\(q_{Sn} = (229 g)(0.213 J/g*°C)(26.4 °C - 110 °C) = -4.077*10^{3} J\)
\(q_{H_{2}O} = (100g)(4.184 J/g*°C)(26.4 °C - 21.0 °C) = 2.259*10^{3} J\)
\(q_{cal}+q_{Sn}+q_{H_{2}O}=0\)
\(q_{cal} = -(q_{Sn}+q_{H_{2}O})\)
qcal = -(-4.077*103 J + 2.259*103 J) = 1.818*103J
\(q_{cal}=C_{c}\Delta T_{c}\)
1.82*103 J = Cc (26.4-21.0)
Cc = 336.7 J/°C
Exercise \(\PageIndex{4}\)
A chemical reaction in a bomb calorimeter evolves 4.12 kJ of energy in the form of heat. If the temperature of the bomb calorimeter increases by 5.90 K, what is the heat capacity of the calorimeter?
- Answer
-
698 J/K
\(\Delta T = 5.90 K\)
\(q_{cal} = 4.12 kJ\)
\(C = \frac{q_{cal}}{\Delta T} = \frac{4.12 kJ}{5.90 K} = 0.6983 kJ/K\)