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Wize University Physics Textbook (Master) > Heat and Temperature

Calorimetry

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Specific Heat & Calorimetry

  • Heat required to change the temperature of mass mm by ΔT\Delta T is:
Q=mcΔTQ=mc\Delta T
  • cc is called specific heat of the material which is the heat required to change the temperature of 1 kg of a material by 1 degree of Kelvin
  • Q>0Q > 0 means that heat is added to the material and Q<0Q < 0 means that heat is removed from the material
MaterialC[Jkg⋅K]Aluminum910Copper390Ice(near0°C)2100Iron470Water(liquid)4190\begin{array}{|l|c|}\hline Material & C [\dfrac{J}{kg\cdot K}]\\[5pt]\hline Aluminum & 910\\[5pt]\hline Copper & 390\\[5pt]\hline Ice (near 0\degree C) & 2100\\[5pt]\hline Iron & 470\\[5pt]\hline Water (liquid) & 4190\\[5pt]\hline \end{array}
Specific Heat For Different Material



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Phase Change

  • The heat required to change the phase of the material at a constant temperature is:
Q=±mLQ=\pm mL
  • Where LL in this equation could be:
  • LfL_f (heat of fusion) for melting
  • LvL_v (heat of vaporization) for vaporization
  • Phase change happens at constant temperature!
  • Phase change is a reversible process
  • During a phase change both phases co-exist!
  • For example: mixture of ice and water at 0°C0\degree C
SubstanceLf[Jkg]Lv[Jkg]Water334×1032256×103Copper134×1035069×103Ethanol104.2×103854×103Nitrogen25/5×103201×103\begin{array}{|l|c|c|}\hline \\[-3pt]Substance & L_f [\frac{J}{kg}] & L_v [\frac{J}{kg}]\\[7pt]\hline \\[-3pt]Water & 334 \times10^3 & 2256 \times10^3\\[5pt]\hline \\[-3pt]Copper & 134 \times10^3 & 5069 \times10^3\\[5pt]\hline \\[-3pt]Ethanol & 104.2 \times10^3 & 854 \times10^3\\[5pt]\hline \\[-3pt]Nitrogen & 25/5 \times10^3 & 201 \times10^3\\[5pt]\hline \end{array}
Heat of Fusion & Vaporization for Common Substances
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Example

Below is a phase change diagram of a material by adding heat to it at a constant rate


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Example: Calorimetry

An unknown mass of water at temperature of 100 ºC is added to 500 cm3 of water at 15 ºC in a calorimeter. If eventually there is water at 75 ºC in the calorimeter, what is the mass of water at 100ºC added initially? (Cw=4.179J/g0CC_w=4.179 J/g^0 C)

Solution:

Vcw=500cm3ρw=1g/cm3⇒ρmV⇒mcw=500gV_{cw}=500cm^3\hspace{0.5cm}\rho_w=1g/cm^3\Rightarrow \rho \dfrac{m}{V}\Rightarrow m_{cw}=500g
Ti−cw=15°CT_{i-cw}=15\degree C
Tf−cw=75°CT_{f-cw}=75\degree C
ΔTcw=60°C\Delta T_{cw}=60\degree C
ΔThw=(75°C−100°C)\Delta T_{hw}=(75\degree C-100\degree C)
=−25°C=-25\degree C
mhw=?m_{hw}=?
ρw=1g/cm3⇒ρ=mv⇒mcw=500 g\rho_w=1g/cm^3\quad\Rightarrow\rho=\frac{m}{v}\quad\Rightarrow m_{cw}=500\ g
cw=4.179J/g°Cc_w=4.179J/g\degree C


−Q1=Q2-Q_1=Q_2

−mhwcwΔThw=mcwcwΔTcw-m_{hw}c_w\Delta T_{hw}=m_{cw}c_w\Delta T_{cw}

mhw=−mcwΔTcwΔThwm_{hw}=\dfrac{-m_{cw}\Delta T_{cw}}{\Delta T_{hw}}

mhw=−(500g)(60°C)(−25°C)=1200gm_{hw}=\frac{-(500g)(60\degree C)}{(-25\degree C)}=1200g
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Example: Phase Change

A 0.60 kg piece of ice at 0ºC is submerged in a cup of water with 250ml of water at room temperature. Assuming that the system is isolated and no heat is lost to the environment, what is the final temperature of the water? (LF = 0.80 kcal/kg and Troom=21.5°CT_{room}=21.5 \degree C, Cwater=1.0kcal/kgCC_{water}=1.0kcal/kgC)

Solution:

mice=0.60kgm_{ice}=0.60kg
Tw=21.5°CT_w=21.5\degree C
m=0.250kgm=0.250kg
Tice=0°Cbegin to meltT_{ice}=0\degree C\hspace{0.2cm}\text{begin to melt}

1 cup=0.250L1\ cup=0.250L

⇒250mL=250cm3\Rightarrow 250mL=250cm^3

ρ=1g/cm3\rho = 1g/cm^3

=0.250kg=0.250kg

Q1=miceLfusionQ2=miceCwaterΔTQ3=MwaterCwaterΔTQ_1=m_{ice}L_{fusion}\hspace{0.3cm}Q_2=m_{ice}C_{water}\Delta T\hspace{0.3cm}Q_3=M_{water}C_{water}\Delta T

miceLfusion+miceCw(Tf−0°)=−mwcw(Tf−21.5°C)m_{ice}L_{fusion}+m_{ice}C_w{(T_f-0\degree })=-m_wc_w(T_f-21.5\degree C)

miceLf+micecwTf+mwcwTf=mwcw(21.5°C)m_{ice}L_f+m_{ice}c_wT_f+m_wc_wT_f=m_wc_w(21.5\degree C)

Tf=mwcw(21.5°C)−miceLfmicecw+mwcwT_f=\dfrac{m_wc_w(21.5\degree C)-m_{ice}L_f}{m_{ice}c_w+m_wc_w}

=(0.250kg)(1.0kcal/kgC)(21.5oC)−(0.60kg)(0.80)((0.60kg)(1.0)+(0.250kg)(1.0))= \dfrac{(0.250kg)(1.0kcal/kgC)(21.5^oC)-(0.60kg)(0.80)}{\left( (0.60kg)(1.0)+(0.250kg)(1.0)\right)}

=5.76°C=5.76\degree C

Tf=5.76°CT_f=5.76\degree C
Answer the following questions:
Part 1
If we add one 20 g20\ g ice cube at −10°C−10\degree C into a glass with 100 g100\ g of water at 20°C20\degree C, what is the final temperature at thermal equilibrium? (in degrees of Celcius)
Cw=4190JKg⋅K, Ci=2100JKg⋅K, Li=334×103JkgC_w=4190\dfrac{J}{Kg\cdot K},\ C_i=2100\dfrac{J}{Kg\cdot K},\ L_i=334\times10^3\dfrac{J}{kg}

Extra Practice
Calorimetry
Answer the following questions:
Practice: Calorimetry Part 1
Practice: Calorimetry
Answer the following questions.
A solution of ice water that weighs 1 kg1\ kg is mixed with 1 kg1\ kg of water at 100°C100\degree C. After mixing the temperature of the final solution is at 24°C24\degree C. How much of the ice water solution was ice.
Latent heat of ice: 334 Jg334\ \frac{J}{g}
C value for water: 4.186 Jg4.186\ \frac{J}{g}
Liquid helium is stored at its boiling point, 4.2 K4.2\ K, in a spherical capsule that is separated by a vacuum space from a surrounding container that is maintained at the temperature of liquid nitrogen, 77 K77\ K. The capsule has a radius of 0.1 m0.1\ m and it is blackened on the outside.
a) Calculate the amount of helium that boils away in one day.
b) Now imagine the vacuum is replaced by a 20 cm20\ cm glass layer around the spherical capsule. How much helium boils away per day in this new condition?
A copper sphere with a radius of 5 cm5\ cm at T=80°CT=80\degree C is placed into a 2 kg2\ kg steel vessel which is full of water and ice.
What is the final temperature of the system if the final radius of the copper sphere is 4.996 cm4.996\ cm?
If 1/3 of the initial ice-water mixture in the steel vessel is water, how much ice is melted in this process?(gr)
A compound bar made up of 15 cm15\ cm of Cu with Acu=25 cm2A_{cu}=25\ cm^2 and 20 cm20\ cm of steel with ASt=10 cm2A_{St}=10\ cm^2 and T=293 KT=293\ K is fixed at both ends as shown in the diagram below. How much is the stress in each metal if the system is heated uniformly by a 700 watt700\ watt heater for 5 minutes?
CCu=390 Jkg⋅KCSt=470 Jkg⋅KYCu=11×1010 Nm2YSt=20×1010 Nm2αCu=1.7×10−51KαSt=1.2×10−51KρCu=9 kg/cm3ρSt=7.85 kg/cm3\begin{aligned} C_{Cu}=390\ \frac{J}{kg\cdot K} && C_{St}=470\ \frac{J}{kg\cdot K}\\ Y_{Cu}=11\times10^{10}\ \frac{N}{m^2} && Y_{St}=20\times10^{10}\ \frac{N}{m^2}\\ \alpha_{Cu}=1.7\times10^{-5} \frac{1}{K} && \alpha_{St}=1.2 \times10^{-5} \frac{1}{K}\\ \rho_{Cu}=9\ kg/cm^3 && \rho_{St}=7.85\ kg/cm^3 \end{aligned}
Two liters of juice at 20ºC is put in the fridge to cool. How much energy must be removed from the juice to cool it to 8ºC?
When 500g of copper (c=4190 J/kg kc=4190\ J/kg\ k), at 80ºC is added to 500g of water at 10ºC, what is the resulting temperature of the mixture?
Two ice cubes, each of 3.0g are added to lukewarm bottle of water at 7ºC. How much of the ice’s mass will melt? (Lf=333×103 j/kg,L_f=333\times10^3\ j/kg, and 4190 j/kg k4190\ j/kg \ k).
Calorimetry
Answer the following questions:
Practice: Calorimetry Part 2
Part 2
What happens if we add two 20 g20\ g ice cubes at −10°C-10\degree C into the same glass we had in part I. What is temperature at equilibrium? (temperature is in degrees of Celcius)
Cw=4190JKg⋅K, Ci=2100JKg⋅K, Li=334×103JkgC_w=4190\dfrac{J}{Kg\cdot K},\ C_i=2100\dfrac{J}{Kg\cdot K},\ L_i=334\times10^3\dfrac{J}{kg}
The following diagram shows the heating process of 200 g200\ g of an unknown substance at constant rate. The substance is solid at 20°C20\degree C. (Csolid=1300Jkg⋅KC_{solid}=1300 \frac{J}{kg\cdot K})
What is the power of heating?
What is the heat of fusion of this substance?
Practice: Calorimetry Part 3
Part 3
How much ice is left in the glass when the system in part 2 is at thermal equilibrium?
Cw=4190JKg⋅K, Ci=2100JKg⋅K, Li=334×103JkgC_w=4190\dfrac{J}{Kg\cdot K},\ C_i=2100\dfrac{J}{Kg\cdot K},\ L_i=334\times10^3\dfrac{J}{kg}
New Practice Question
A copper sphere with a radius of 5𝑐𝑚 𝑎𝑡 𝑇 = 80℃ is placed into a 2 kg steel vessel which is full of water and ice.
a. What is the final temperature of the system if the final radius of the
copper sphere is 4.996cm?
A copper rod in the form of a cylinder of radius 1cm and length of 1m is heated to 600°C and dropped into 2.00 kg of water at 20.0°C. Assuming that no energy is lost by heat to the surroundings, determine the final equilibrium temperature of the system.
density of Cu = 8.94g/cm3 c(Cu)=385 J/(kgC) c(water)=4182 J/(kgC)