An online conversion calculator for the units of energy Watt-hour (Wh) and Joules (J) is presented. Examples and problems involving the conversion of Watt-hours and Joules are also included.
The Watt-hour (Wh) and Joules (J) are units of energy with the conversion rate as follows:
\[ 1 \text{ Watt-hour (Wh)} = 3600 \text{ Joules (J)} \] \[ 1 \text{ Joule (J)} = \left(\frac{1}{3600}\right) \text{ Watt-hour (Wh)} \]Reference: NIST Reference Guidelines
Convert \( 15.6 \, \text{Watt-hour (Wh)} \) to Joules (J).
Given that \( 1 \, \text{Wh} = 3600 \, \text{J} \):
\[ 15.6 \, \text{Wh} = 15.6 \times 3600 \, \text{J} = 56,160 \, \text{J} \]Convert \( 5,660 \, \text{Joules} \) to Watt-hour (Wh) and round to the nearest tenth of a Wh.
Given that \( 1 \, \text{J} = \left(\frac{1}{3600}\right) \text{Wh} \):
\[ 5,660 \, \text{J} = 5,660 \times \left(\frac{1}{3600}\right) \text{Wh} \approx 1.57222 \, \text{Wh} \]Rounding to the nearest tenth of a Wh:
\[ 5,660 \, \text{J} = 1.6 \, \text{Wh} \]Enter the number of Watt-hours (Wh) or the number of Joules (J) to perform conversions.
The cost of \( 1,000 \, \text{Wh} \) of electricity is 25 cents. How much does \( 10^6 \, \text{Joules} \) of electricity cost? Round the answer to the nearest cent.
Given that \( 1 \, \text{Wh} = 3600 \, \text{J} \):
\[ 1,000 \, \text{Wh} = 1,000 \times 3600 \, \text{J} = 3,600,000 \, \text{J} \]The cost of \( 3,600,000 \, \text{J} \) is 25 cents. The cost \( C \) of \( 10^6 \, \text{J} \) is given by:
\[ C = \left(\frac{25 \, \text{cents}}{3,600,000 \, \text{J}}\right) \times 10^6 = 6.9444 \, \text{cents} \]Rounding to the nearest cent:
\[ C = 7 \text{ cents} \]It takes \( 4,180 \, \text{Joules} \) to raise the temperature of \( 1 \, \text{Kg} \) of water by \( 1^\circ\text{C} \). How many Watt-hours are needed to raise the temperature of \( 1 \, \text{Kg} \) of water by \( 1^\circ\text{C} \)? Round to the nearest tenth.
Given that \( 1 \, \text{J} = \left(\frac{1}{3600}\right) \text{Wh} \):
\[ 4,180 \, \text{J} = 4,180 \times \left(\frac{1}{3600}\right) \text{Wh} \approx 1.16111 \, \text{Wh} \]Rounding to the nearest tenth:
\[ 1.2 \, \text{Wh} \]One cubic foot of natural gas can produce about \( 10^6 \, \text{Joules} \) of energy. How many Watt-hours may be produced by 1 cubic foot of natural gas? Round to the nearest unit.
Given that \( 1 \, \text{J} = \left(\frac{1}{3600}\right) \text{Wh} \):
\[ 10^6 \, \text{Joules} = 10^6 \times \left(\frac{1}{3600}\right) \text{Wh} \approx 277.7778 \, \text{Wh} \]Rounding to the nearest unit:
\[ 278 \, \text{Wh} \]The specific heat capacity of water is \( 4180 \, \text{J/(kg}\cdot^\circ\text{C)} \), which means you need \( 4180 \, \text{J} \) to raise the temperature of \( 1 \, \text{Kg} \) of water by \( 1^\circ\text{C} \). A domestic solar panel system produces on average \( 20,000 \, \text{Wh} \) of electricity per day. How many Kg of water are heated from \( 40^\circ\text{C} \) to \( 100^\circ\text{C} \) per day by this system? Round to the nearest Kg.
Hint: The formula that gives the energy \( E \) needed to raise \( m \) kg of water from temperature \( T_1 \) to \( T_2 \) is given by: \( E = m \times (4180 \, \text{J/(kg}\cdot^\circ\text{C)}) \times (T_2 - T_1) \).
First, convert the energy produced per day from Wh to J (since \( 1 \, \text{Wh} = 3600 \, \text{J} \)):
\[ 20,000 \, \text{Wh} = 20,000 \times 3600 \, \text{J} = 7.2 \times 10^7 \, \text{J} \]Let \( m \) be the mass of water in Kg. Using the energy formula:
\[ 7.2 \times 10^7 = m \times (100 - 40) \times 4180 = m \times 60 \times 4180 = m \times 250,800 \]Solve for \( m \):
\[ m = \dfrac{7.2 \times 10^7}{250,800} \approx 287.0813 \, \text{Kg} \]Rounding to the nearest Kg:
\[ m = 287 \, \text{Kg} \]