Chapter 14, Problem 14P

A 31.5-g glass thermometer reads 23.6°C before it is placed in 135 mL of water. When the water and thermometer come to equilibrium, the thermometer reads 41 8°C. What was the original temperature of the water? Ignore the mass of fluid inside the glass thermometer.

To determine

Equilibrium temperature of a thermometer

Answer to Problem 14P

Solution:

Equilibrium temperature of a thermometer is 42.6 C.

Explanation of Solution

Given:

$cpa$ is the specific heat of glass which is 792 J/kgK

$cpb$ is the specific heat of water which is 4186 J/kgK

$ma$ is the mass of glass which is 31.5 g

$mb$ is the mass of water

$\rho$ is density of water is 1000 kg/cu.m

$V$ is volume which is 135 mL or 0.135 L

$Ta$ is the temperature of glass which is 23.6 C or 296.6 K

$Tb$ is the temperature of water

$Te$ is equilibrium temperature which is 41.8 C or 314.8 K

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 can be derived.

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

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

$Qa=macpa(Te-Ta)$

$Qb=mbcpb(Te-Tb)$

Equating the heat transferred is $Qa=-Qb$

This becomes

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

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

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

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

Where,

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

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

$ma$ is the mass of material $a$

$mb$ is the mass of material $b$

$Ta$ is the temperature of material $a$

$Tb$ is the temperature of material $b$

$Te$ is equilibrium temperature

Calculation:

The mass of water $mb$ is determined by the multiplying density $\rho$ with volume $V$

$mb=\rho V$

Volume $V$ by the multiplying is 0.135 L. 1000 L is 1 cu.m. So 0.135 L is 0.000135 cu.m.

Mass is now

$mb=(1000)(0.000135)$

$mb=0.135$ kg

Inserting the given values in the equilibrium temperature formula

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

$314.8=\frac{(0.0315)(792)(296.6)+(0.135)(4186)(Tb)}{(0.0315)(792)+(0.135)(4186)}$

$Tb=315.6$ K

$Tb=42.6$ C