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A copper sphere with a radius of 5 cm at T=80 C is placed into a 2 kg steel ves…

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Wize University Physics Textbook (Master) > Heat and Temperature

Calorimetry

4 Activities

Wize University Physics Textbook (Master) > Heat and Temperature

Thermal Stress

3 Activities

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.
  1. What is the final temperature of the system if the final radius of the copper sphere is 4.996 cm4.996\ cm?
  2. If 1/3 of the initial ice-water mixture in the steel vessel is water, how much ice is melted in this process?(gr)
Cw=4190Jkg⋅KLf=334×103Jkg⋅KCSt=470Jkg⋅KCcu=390Jkg⋅Kρcu=9g/cm3αCu=1.7×10−51K\begin{aligned} C_w=4190 \frac{J}{kg\cdot K} &&& L_f=334\times10^3\frac{J}{kg\cdot K}\\ C_{St}=470 \frac{J}{kg\cdot K}&&& C_{cu}=390\frac{J}{kg\cdot K}\\ \rho_{cu}=9 g/cm^3 &&& \alpha_{Cu}=1.7\times10^{-5} \frac{1}{K} \end{aligned}
Question A
More Calorimetry Questions:
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 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)
More Thermal Stress Questions:
Practice: Thermal Stress
An aluminum rod is attached to a steel bar as shown in the figure below so that they expand or stretch together. The length of these bars is 2 m2\ m at 10°C10\degree C and the cross-sectional area of both bars are the same. How much is the stress in each bar at 100°C100\degree C?
αSt=1.2×10−51KYSt=20.0×1010Nm2αAl=2.4×10−51KYAl=69×109Nm2\begin{array}{ll} \alpha_{St}=1.2\times10^{-5} \frac{1}{K} & Y_{St}=20.0\times10^{10} \frac{N}{m^2}\\[10pt] \alpha_{Al}=2.4\times10^{-5} \frac{1}{K} & Y_{Al}=69\times10^9 \frac{N}{m^2} \end{array}
Practice: Thermal Stress
An aluminum rod is attached to a steel bar as shown in the figure below so that they expand or stretch together. The length of these bars is 2 m2\ m at 10°C10\degree C and the cross-sectional area of both bars are the same. How much is the stress in each bar at 100°C100\degree C?
αSt=1.2×10−51KYSt=20.0×1010Nm2αAl=2.4×10−51KYAl=69×109Nm2\begin{array}{ll} \alpha_{St}=1.2\times10^{-5} \frac{1}{K} & Y_{St}=20.0\times10^{10} \frac{N}{m^2}\\[10pt] \alpha_{Al}=2.4\times10^{-5} \frac{1}{K} & Y_{Al}=69\times10^9 \frac{N}{m^2} \end{array}
When making cars companies cool aluminum bolts to −50° C-50\degree\ C so that they perfectly fit into the car when inserted. After being inserted into the car the bolts warm up to room temperature which is 20° C20\degree\ C.
Find the Stress caused from the bolt after it heats up to room temperature.
αAl=22⋅10−6 K−1α_{Al}=22\cdot10^{-6}\ K^{-1}
A 3 m3\ m bronze bar with a cross-sectional area of 2.5 cm22.5\ cm^2 is placed between two rigid walls as shown below. At T=−15°CT=-15\degree C, the gap is Δ=2 mm\Delta =2\ mm What temperature will make the stress in the bar to be 40 MPa40\ ΜPa?
(αbr=18.0×10−61K\alpha_{br}=18.0\times10^{-6} \frac{1}{K} and Ybr=80×109 PaY_{br}=80\times10^9\ Pa)
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}
New Practice Question
At 80 degrees of Celsius, a steel tire 10 mm thick and 80 mm wide that is to be shrunk onto a locomotive driving wheel 2 m in diameter just fits over the wheel which is at temperature of 25°C. How much is the stress on the tire when the combination cools down to 25°C? Neglect the deformation of the wheel.
(αSt=11.7×10−6 1K and YSt=200×109 Pa)\left(\alpha_{St}=11.7\times10^{-6}\ \frac{1}{K}\ and\ Y_{St}=200\times10^9\ Pa\right)
Input your answer in Pa, to 4 sig. figs., do NOT include units.
New Practice Question
An aluminum rod is attached to a steel bar as shown in the figure below so that they expand or stretch together. The length of these bars is 2𝑚 at 10℃ and the cross-sectional area of both bars is the same. How much is the stress in each bar at 100℃?
αSt=1.2×10−5 1K          YSt=20.0×1010 Nm2\alpha_{St}=1.2\times10^{-5}\ \frac{1}{K}\ \ \ \ \ \ \ \ \ \ Y_{St}=20.0\times10^{10}\ \frac{N}{m^2}
αAl=2.4×10−5 1K          YAl=69.0×109 Nm2\alpha_{Al}=2.4\times10^{-5}\ \frac{1}{K}\ \ \ \ \ \ \ \ \ \ Y_{Al}=69.0\times10^9\ \frac{N}{m^2}
Thermal Stress
An aluminum rod is attached to a steel bar as shown in the figure below so that they expand or stretch together. The length of these bars is 2 m2\ m at 10°C10\degree C and the cross-sectional area of both bars are the same. How much is the stress in each bar at 100°C100\degree C?
αSt=1.2×10−51KYSt=20.0×1010Nm2αAl=2.4×10−51KYAl=69×109Nm2\begin{array}{ll} \alpha_{St}=1.2\times10^{-5} \frac{1}{K} & Y_{St}=20.0\times10^{10} \frac{N}{m^2}\\[10pt] \alpha_{Al}=2.4\times10^{-5} \frac{1}{K} & Y_{Al}=69\times10^9 \frac{N}{m^2} \end{array}
Thermal Expansion
How much taller does the Eiffel Tower become at the end of a day when the temperature has increased by 15oC. Its original height is 321m and you can assume it is made of steel.
Thermal Stress
An Aluminium rod of length of 4 meters and a cross sectional area of 10 cm210\ cm^2 weighs 1 kg. It is secured between two stable supports. There is no initial stress in the beam. If the temperature is initially 60 degrees and increases by 40 degrees. What is the stress in the beam. How much water at 20 degrees C must the 100 degree beam be put in to come to 50 degrees C.
αAl=22⋅10−6 K−1α_{Al}=22\cdot10^{-6}\ K^{-1}
YAl=65⋅109 PaY_{Al}=65\cdot10^9\ Pa
An Aluminium rod of length of 4 meters and a cross sectional area of 10 cm210\ cm^2. It is secured between two stable supports. There is an initial stress in the beam of 5⋅107 Nm25\cdot10^7\ \frac{N}{m^2}. If the temperature increases by 30 degrees. What is the new stress in the beam.
αAl=22⋅10−6 K−1α_{Al}=22\cdot10^{-6}\ K^{-1}
YAl=65⋅109 PaY_{Al}=65\cdot10^9\ Pa
Thermal Stress
We are planning to close a hole using an aluminum ball but the ball is too small to cover the entire hole. What could we do to make it fit the hole?