Chapter 14, Problem 13P

To determine

Equilibrium temperature of copper in an aluminium and water calorimeter.

Answer to Problem 13P

Solution:

Equilibrium temperature of copper in an aluminium and water calorimeter is 75.3 C

Explanation of Solution

Given:

$cpa$ is the specific heat of copper which is 376 J/kgk

$cpb$ is the specific heat of aluminium which is 921 J/kgk

$cpc$ is the specific heat of water which is 4186 J/kgk

$ma$ is the mass of copper which is 265 g or 0.265 kg

$mb$ is the mass of aluminium which is 145 g

$mc$ is the mass of water which is 825 g or 0.825 kg

$Ta$ is the temperature of copper which is 245 C

$Tb$ is the temperature of aluminium which is 12 C

$Tc$ is the temperature of water which is 12 C

Formula Used:

The formula used is the relation between temperature and specific heat from thermodynamics

$Q=mcp(Tf-Ti)$

Where,

$Q$ is the heat transferred

$cp$ is the specific heat

$m$ is the mass

$Tf$ is the final temperature

$Ti$ is the initial temperature

From the formula above, the equilibrium temperature formula for calorimetry can be derived.

If two materials $a$, $b$ and calorimeter $c$ with temperatures $Ta$, $Tb$ and $Tc$ are in contact, then they will reach an equilibrium temperature $Te$.

This equilibrium temperature $Te$ is given by equating the heat transferred by the three materials.

$Qa=macpa(Te-Ta)$

$Qb=mbcpb(Te-Tb)$

$Qc=mccpc(Te-Tc)$

Equating the heat transferred is $Qa=-(Qb+Qc)$

This becomes

$Qa=macpa(Te-Ta)=-Qb-Qc=-mbcpb(Te-Tb)-mccpc(Te-Tc)$

$macpa(Te-Ta)=-mbcpb(Te-Tb)-mccpc(Te-Tc)$

$macpa(Te-Ta)=mbcpb(Tb-Te)+mccpc(Tc-Te)$

$Te=\frac{macpa(Ta)+mbcpb(Tb)+mccpc(Tc)}{macpa+mbcpb+mccpc}$

Where,

$cpa$ is the specific heat of material $a$

$cpb$ is the specific heat of material $b$

$cpc$ is the specific heat of material $c$

$ma$ is the mass of material $a$

$mb$ is the mass of material $b$

$mc$ is the mass of material $c$

$Ta$ is the temperature of material $a$

$Tb$ is the temperature of material $b$

$Tc$ is the temperature of material $c$

$Te$ is equilibrium temperature

Calculation:

Inserting the given values in the equilibrium temperature formula for calorimetry:

$Te=\frac{macpa(Ta)+mbcpb(Tb)+mccpc(Tc)}{macpa+mbcpb+mccpc}$

$Te=\frac{(0.265)(376)(245)+(0.145)(921)(12)+(0.825)(4186)(12)}{(0.265)(376)+(0.145)(921)+(0.825)(4186)}$

$Te=75.3$ C