Chapter 14, Problem 11P

(a) How much energy is required to bring a 1.0-L pot of water at 20 C to 100 C? (b) For how long could this amount of energy run a 60-W lightbulb?

To determine

Part(a)To determine:

Energy required to boil a litre of water

Answer to Problem 11P

Solution:

Energy required to boil a litre of water is 334880 J

Explanation of Solution

Given:

$Q$ is to be determined

$cp$ of water is 4186 J/kgK

$m$ is mass

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

$V$ is volume which is 1 L

$Tf$ is 100 C

$Ti$ is 20 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

Calculation:

The mass of water $m$ is determined by the multiplying density $\rho$

with volume $V$

$m=\rho V$

Volume $V$

by the multiplying is 1 L. 1000 L is 1 cu.m. So 1 L is 0.001 cu.m.

Mass is now

$m=(1000)(0.001)$

$m=1$ kg

Inserting the given values in the heat capacity formula: $Q=mcp(Tf-Ti)$

$Q=(1)(4186)(100-20)$

$Q=\text{334880}$ J

To determine

Part(b)To determine:

The duration a light bulb will last for a given energy.

Answer to Problem 11P

Solution:

The duration a light bulb will last is 5581 seconds.

Explanation of Solution

Given:

$Q$ is 334880 J from part(a)

$P$ is 60 W

Formula Used:

Formula for power

$P=\frac{Q}{t}$

Where,

$Q$ is the heat transferred

$P$ is Power

$t$ is time

Calculation:

An energy source of energy $Q$ will run the light bulb for a certain time. If power of the bulb is known then one can calculate the time.

The formula is $P=\frac{Q}{t}$

$Q$ is 334880 J from part(a)

$P$ is 60 W

$P=\frac{Q}{t}$

$t=\frac{Q}{P}$

Substituting the values above

$t=\frac{334880}{60}$

$t=5581$ seconds