Chapter 14, Problem 16P

To determine

The specific heat of glycerine.

Answer to Problem 16P

Solution:

The specific heat of glycerineis2356 J/kgK.

Explanation of Solution

Given:

$cpa$ is the specific heat of iron which is 460 J/kgk

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

$cpc$ is the specific heat of glycerine

$ma$ is the mass of iron which is 290 g or 0.290 kg

$mb$ is the mass of aluminium which is 95 g or 0.095 kg

$mc$ is the mass of glycerine which is 250 g or 0.250 kg

$Ta$ is the temperature of copper which is 180 C

$Tb$ is the temperature of aluminium which is 10 C

$Tc$ is the temperature of glycerine which is 10 C

$Te$ is equilibrium temperature which is 38 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:

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

$38=\frac{(0.290)(460)(180)+(0.095)(921)(10)+(0.250)(cpc)(10)}{(0.290)(460)+(0.095)(921)+(0.250)(cpc)}$

$cpc=2356$ J/kgK