Q.3.27 - An Introduction to General, Organic, and Biological Chemistry

Question

The energy needed to keep a 75 -watt light bulb burning for 1hr is 270 kJ. Calculate the energy required to keep the light bulb burning for 3h in each of the following energy units:

a. joules

b. kilocalories

Step-by-Step Solution

Answer

(Part a) $8.1\times {10}^5\mathrm{J}$

(part b) $193.6 kcal$

Part (a) Step 1: Given Information

The energy needed to keep a $75$ -watt light bulb burning for one hour is $270 kJ$ 

We have to find out the energy required to keep the light bulb burning for $3$hr in the unit joules.

(Part a) Step 2: Explanation

Given,

$270 kJ$ is required for an $75$ watt light bulb to burn for one hour

Hence the energy required to keep the light bulb burning for $3hr$in joules is,

$1 \mathrm{kJ}=1000 \mathrm{J}$

$$\begin{gathered} =3\times 270=810 kJ \\ =8.1\times {10}^{5}J \end{gathered}$$

Part (b) Step 3: Given Information

The energy needed to keep a$75$-watt light bulb burning for one hour is$270 kJ$

We have to find the energy required to keep the light bulb burning for $3h$ in the unit kilocalories.

Part (b) Step 4: Explanation

Given,

$270 kJ$ is required for an $75$ watt light bulb to burn for one hour

Hence the energy required to keep the light bulb burning for $3hr$ in kilocalories is,

$1 \text{k}\mathrm{cal}=4.184 \mathrm{kJ}$

$$\begin{gathered} \Rightarrow \frac{810}{4.184} \\ =193.6 kcal \end{gathered}$$