Calculating Energy Stores
Key Stage 4
Meaning
Energy stores can be calculated from other known facts about an object.
About Calculating Energy Stores
- Calculating energy stores can be done using information about properties, distances, motion and fields affecting an object.
The following energy stores can be calculated from other quantities:
- Thermal Energy = (Mass) x (Specific Heat Capacity) x (Change in Temperature)
- Elastic Potential Energy = 0.5 x (Spring Constant) x (Extension)2
- Kinetic Energy = 0.5 x (Mass) x (Speed)2
- Gravitational Potential Energy = (Mass) x (gravitational field strength) x (change in height)
Examples
| A 2 kg object made from a material with specific heat capacity 5 J/kg/°C is heated by 10°C. Calculate the Thermal Energy transferred to the object. | A sealed metal can containing 5.5 kg of water is heated over a fire from 27°C to 100°C to sterilise the water. Calculate the increase in thermal energy of the water.
The Specific Heat Capacity of water is 4200 J/kg/°C. (Give your answer correct to 2 significant figures.) |
A 0.5kg Iron baking tray is heated from 20°C to 80°C in an oven. Calculate the work done by the oven in heating the tray.
The Specific Heat Capacity of Iron is 450 J/kg/°C. (Give your answer correct to 2 significant figures.) |
| 1. State the known quantities
m = 2 kg c = 5 J/kg/°C Δθ = 10°C |
1. State the known quantities
m = 5.5 kg c = 4200 J/kg/°C θ1 = 27°C θ2 = 100°C Find the temperature change. Δθ = θ2 - θ1 = 100 - 27 = 73°C |
1. State the known quantities
m = 0.5 kg c = 450 J/kg/°C θ1 = 20°C θ2 = 80°C Find the temperature change. Δθ = θ2 - θ1 = 80 - 20 = 60°C |
| 2. Substitute the numbers into the equation and solve.
\(E_t = m c \Delta \theta\) \(E_t = 2 \times 5 \times 10\) \(E_t = 100J\) |
2. Substitute the numbers into the equation and solve.
\(E_t = m c \Delta \theta\) \(E_t = 5.5 \times 4200 \times 73\) \(E_t = 1686300J\) \(E_t \approx 1700000J\) |
2. Substitute the numbers into the equation and solve.
\(E_t = m c \Delta \theta\) \(E_t = 0.5 \times 450 \times 60\) \(E_t = 13500J\) \(E_t \approx 14000J\) |
| A bow with a spring constant of 400N/m is stretched 0.5m with a force of 200N. Calculate the elastic potential store of the bow. | A bungee cord with a spring constant of 45N/m stretches by 30m. Calculate the elastic potential store of the cord.
Give your answer correct to 2 significant figures. |
A slinky spring of length 0.1m and spring constant 0.80N/m is stretched to a length of 9.1m. Calculate the elastic potential store of the slinky.
Give your answer correct to 2 significant figures. |
| 1. State the known quantities
k = 400N/m x = 0.5m |
1. State the known quantities
k = 45N/m x = 30m |
1. State the known quantities
k = 0.8N/m Original Length = 0.1m Final Length = 9.1m Find the extension. x = Final Length - Original Length = 9.1 - 0.1 = 9.0m |
| 2. Substitute the numbers into the equation and solve.
\(E_e = \frac{1}{2} k x^2\) \(E_e = \frac{1}{2} \times 400 \times 0.5^2\) \(E_e = \frac{1}{2} \times 400 \times 0.25\) \(E_e = 50J\) |
2. Substitute the numbers into the equation and solve.
\(E_e = \frac{1}{2} k x^2\) \(E_e = \frac{1}{2} \times 45 \times 30^2\) \(E_e = \frac{1}{2} \times 45 \times 900\) \(E_e = 20250J\) \(E_e \approx 20,000J\) |
2. Substitute the numbers into the equation and solve.
\(E_e = \frac{1}{2} k x^2\) \(E_e = \frac{1}{2} \times 0.80 \times 9.0^2\) \(E_e = \frac{1}{2} \times 0.80 \times 81\) \(E_e = 32.4J\) \(E_e \approx 32J\) |
| A 700kg formula one racing car has a top speed of 100m/s. Calculate the kinetic energy of the car to two significant figures. | A cheetah of mass 75kg runs at a speed of 32m/s. Calculate the kinetic energy of the cheetah correct to two significant figures. | A 160g cricket ball is hit at 44m/s. Calculate the kinetic energy of the cricket ball correct to two significant figures. |
| 1. State the known quantities
m = 700kg v = 100m/s |
1. State the known quantities
m = 75kg v = 32m/s |
1. State the known quantities
m = 160g min kilograms = \(\frac{160}{1000}\) m = 0.16kg v = 32m/s |
| 2. Substitute the numbers into the equation and solve.
\(E_k = \frac{1}{2} m v^2\) \(E_k = \frac{1}{2} \times 700 \times 100^2\) \(E_k = \frac{1}{2} \times 700 \times 10,000\) \(E_k = 3,500,000J\) |
2. Substitute the numbers into the equation and solve.
\(E_k = \frac{1}{2} m v^2\) \(E_k = \frac{1}{2} \times 75 \times 32^2\) \(E_k = \frac{1}{2} \times 75 \times 1024\) \(E_k = 38,400J\) \(E_k \approx 38000J\) |
2. Substitute the numbers into the equation and solve.
\(E_k = \frac{1}{2} m v^2\) \(E_k = \frac{1}{2} \times m \times v^2\) \(E_k = \frac{1}{2} \times 0.16 \times 44^2\) \(E_k = \frac{1}{2} \times 0.16 \times 1936\) \(E_k = 154.88J\) \(E_k \approx 150J\) |
| A weight lifter lifts a 50kg mass a distance of 2.0m from the ground. Calculate the increase in gravitational potential energy of the mass.
g on Earth is 9.8N/kg |
A pulley is used to lift a 12 tonne mass 0.80m above the ground. Calculate the change in energy in the gravitational potential store.
g on Earth is 9.8N/kg Give your answer correct to two significant figures. |
During a rock slide a 320kg boulder falls from a height of 1450m to a height of 730m above sea level. Calculate the change in gravitational potential energy.
g on Earth is 9.8N/kg Give your answer correct to two significant figures. |
| 1. State the known quantities
m = 50kg g = 9.8N/kg Δh = 2.0m |
1. State the known quantities
m = 12tonne = 12,000kg g = 9.8N/kg Δh = 0.80m |
1. State the known quantities
m = 320kg g = 9.8 N/kg Δh = h2 - h1 = 1450 - 730 = 720m |
| 2. Substitute the numbers into the equation and solve.
\(E_g = m g \Delta h\) \(E_g = m \times g \times \Delta h\) \(E_g = 50 \times 9.8 \times 2\) \(E_g = 980J\) |
2. Substitute the numbers into the equation and solve.
\(E_g = m g \Delta h\) \(E_g = m \times g \times \Delta h\) \(E_g = 0.80 \times 9.8 \times 12000\) \(E_g = 94080J\) \(E_g \approx 94000J\) |
2. Substitute the numbers into the equation and solve.
\(E_g = m g \Delta h\) \(E_g = m \times g \times \Delta h\) \(E_g = 320 \times 9.8 \times 720\) \(E_g = 2257920J\) \(E_g \approx 2300000J\) |