Chapter 14, Problem 15P

A 0.40-kg iron horseshoe, just forged and very hot (Fig. 14-16), is dropped into 1.25 L of water in a 0.30-kg iron pot initially at 20 0°C. If the final equilibrium temperature is 25.0°C, estimate the initial temperature of the hot horseshoe.

Figure 14-16

To determine

Temperature of horseshoe

Answer to Problem 15P

Solution:

Temperature of horseshoe is 26.46 C.

Explanation of Solution

Given:

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

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

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

$ma$ is the mass of iron horseshoe which is 40 kg

$mb$ is the mass of iron pot which is 0.30 kg

$mc$ is the mass of water

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

$V$ is volume which is 1.25 L

$Ta$ is the temperature of iron horseshoe

$Tb$ is the temperature of iron pot which is 20 C

$Tc$ is the temperature of water which is 20 C

$Te$ is equilibrium temperature which is 25 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.

$Qa=macpa(Te-Ta)$

$Qb=mbcpb(Te-Tb)$

$Qc=mccp\mathrm{If\; two\; materials\; c}(Te-Tc)$

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:

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

$mc=\rho V$

Volume $V$ by the multiplying is 1.25 L. 1000 L is 1 cu.m. So 1.25 L is 0.00125cu.m.

Mass is now

$mc=(1000)(0.00125)$

$mc=1.25$ kg

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

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

$25=\frac{(40)(460)(Ta)+(0.3)(460)(20)+(1.25)(4186)(20)}{(40)(460)+(0.3)(460)+(1.25)(4186)}$

$Ta=26.46$ C