Equal 0.400-kg masses of lead and tin at 60.0^∘ C are placed in 1.00 kg of water at 20.0^∘ C. (

Equal 0.400-kg masses of lead and tin at 60.0^∘ C are placed...

Equal $0.400-\mathrm{kg}$ masses of lead and tin at $60.0^{\circ} \mathrm{C}$ are placed in $1.00 \mathrm{~kg}$ of water at $20.0^{\circ} \mathrm{C}_{.}$ (a) What is the equilibrium temperacure of the system? (b) If an alloy is half lead and half tin by mass, what specific heat would you anticipate for the alloy? (c) How many atoms of tin $N_{\mathrm{sa}}$ are in $0.400 \mathrm{~kg}$ of tin, and how many atoms of lead $N_{\mathrm{P} 1}$ are in $0.400 \mathrm{~kg}$ of lead? (d) Divide the number $N_{\mathrm{Sn}}$ of tin atoms by the number $N_{\mathrm{Pb}}$ of lead atoms and compare this ratio with the specific heat of tin divided by the specific heat of lead. What conclusion can be drawn?

Equal $0.400-\mathrm{kg}$ masses of lead and tin at $60.0^{\circ} \mathrm{C}$ are placed in $1.00 \mathrm{~kg}$ of water at $20.0^{\circ} \mathrm{C}_{.}$
(a) What is the equilibrium temperacure of the system? (b) If an alloy is half lead and half tin by mass, what specific heat would you anticipate for the alloy? (c) How many atoms of tin $N_{\mathrm{sa}}$ are in $0.400 \mathrm{~kg}$ of tin, and how many atoms of lead $N_{\mathrm{P} 1}$ are in $0.400 \mathrm{~kg}$ of lead? (d) Divide the number $N_{\mathrm{Sn}}$ of tin atoms by the number $N_{\mathrm{Pb}}$ of lead atoms and compare this ratio with the specific heat of tin divided by the specific heat of lead. What conclusion can be drawn?

Key Concepts

Comparative Analysis of Atomic Content and Macroscopic Properties

This concept explores the relationship between the number of atoms present in a given mass and the macroscopic thermal properties, such as specific heat capacity. By comparing these ratios, one can gain insight into how atomic-scale interactions contribute to the emergent bulk properties of materials.

Alloy Property Estimation

Alloy property estimation involves predicting the overall thermal or mechanical properties of a mixture based on the properties and proportions of its constituent elements. Typically, weighted averages are used to estimate characteristics such as specific heat capacity, demonstrating how mixing materials results in intermediate behavior relative to the pure components.

Mass-Atom Conversion

Mass-atom conversion involves using the molar mass of a substance and Avogadro's number to convert a given mass into the number of atoms. This process bridges macroscopic measurements with the microscopic scale, enabling the analysis of material properties at the atomic level.

Specific Heat Capacity

Specific heat capacity is the amount of heat required to raise the temperature of a unit mass of a substance by one degree Celsius. It is an intrinsic property of materials, and its value influences how substances respond to heat transfer, making it critical in temperature equilibrium calculations and material comparisons.

Conservation of Energy

This fundamental principle states that energy is neither created nor destroyed in an isolated system. In the context of thermal processes, it implies that the heat lost by hotter bodies must equal the heat gained by cooler ones until equilibrium is reached. This concept underpins calorimetry and heat transfer calculations in thermodynamics.

Calorimetry

Calorimetry is the technique used to measure the amount of heat exchanged in physical or chemical processes. By applying the conservation of energy, calorimetry allows the determination of equilibrium temperatures, the calculation of unknown specific heats, and the study of thermal properties in mixtures or reactions.

Transcript

00:02 This problem covers the concept of the calrometry and the heat loss by lead and the team equals the heat cane by water so we can write the heat loss by lead and teen equals the mass of the lead into the specific heat of the lead plus the mass of tin into the specific heat of tin into the initial temperature the initial temperature minus the equilibrium temperature equals the mass of water into the specific heat of water into the equilibrium temperature minus the initial temperature of water so from this we can write the equilibrium temperature t equals the mass of lead plus the mass of lead into the specific heat of lead plus the mass of tin into the specific heat of tin into the initial temperature plus the mass of water into the specific heat of water into the initial temperature of the water upon the mass of water into the specific heat of water plus the mass of lead into the specific heat of lead plus the mass of tin into the specific heat of tin since the mass of lead and tin are equal so we can replace them with m or we can write the equilibrium temperature t equals m into the specific heat of lead plus the specific heat of tin into the initial temperature plus the mass of water into the specific heat of water into the initial temperature of water upon the mass of water into the specific heat of water plus m into the specific heat of lead plus this specific heat of tin now substitute the value to find the equilibrium temperature so for part a the equilibrium temperature equals the mass of lead and tin is 0 .4 kilogram into the specific heat of the specific heat of lead equals 128 joules per kilogram per degree celsius plus the specific heat of tin that is 227 into the initial temperature of lead and tin that is 60 degrees celsius plus the mass of water and that is the mass of water is 1 kilogram into the specific heat that is 4 ,186 joules per kilogram per degree celsius into the initial temperature of water, that is 20 degrees celsius upon 1 kilogram into 4 ,186 joules per kilogram per degree celsius plus 0 .4 kilogram into 1286 joules per kilogram per degree celsius plus 0 .4 kilogram into 128.

04:00 Plus 227 joules per kilogram per degree celsius or the equilibrium temperature comes out to be 21 .3 degrees celsius now part b the the specific heat of the alloy when when equal mass of lead and tin is mixed is given by the mass of lead m into the specific heat of lead plus the mass of tin into the specific heat of tin upon 2m so by substituting this we can write the specific heat of alloy equals 128 plus 227 upon 2 joules per kilogram per degree celsius or the specific heat of the alloy equals 178 joules per kilogram per degree celsius now part c the number of atoms in tin the number of atoms of 10 equals the mass of the tin upon the molecular mass into the regard number or that is equivalent to 0 .4 kilogramm upon the molecular mass is 0 .1 187 kilogram into the avogadro number that is 6 .022 into 10 raise 23 or the number of atoms of tin equals 2 .03 into 10 raise 24 atoms and the number of lead atoms equals 0 .4 kilogram upon the molecular mass of the lead which is 0 .0 .4 kilogramme upon the molecule.

06:26 0 .2072 kilogram into the vaguerre number that is 6 .022 into 10 raise 23 or the number of lead atoms in 0 .4 kj is 1 .16 into 10 .24 atoms.

06:48 Now part d...

Please give Ace some feedback