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

Short answer

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

(part b) localid="1654282169983" $193.6kcal$

Step-by-step solution

Part (a) Step 1: Given Information 

The energy needed to keep a $75$-watt light bulb burning for one hour is localid="1654282180339" $270kJ$

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,

$270kJ$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,

localid="1654282193908" $1\mathrm{kJ}=1000\mathrm{J}$

localid="1654282200518" $$\begin{gathered} =3\times 270=810kJ \\ =8.1\times {10}^{5}J \end{gathered}$$

Part (b)  Step 3: Given Information 

The energy needed to keep a$75$-watt light bulb burning for one hour islocalid="1654282208574" $270kJ$

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,

$270kJ$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,

localid="1654282247095" role="math" $1\text{k}\mathrm{cal}=4.184\mathrm{kJ}$

localid="1654282229811" $$\begin{gathered} \Rightarrow \frac{810}{4.184} \\ =193.6kcal \end{gathered}$$