Caspian's detector logs a sample losing 2 × 10⁻¹² kg of mass, all of it converted to energy. He wants the number of joules. One multiply answers him.
That 2 × 10⁻¹² kg is the given, and a number of joules is the answer. This lesson runs the conversion both ways: mass in, energy out, and energy in, mass out. The equation and the value of c are always given in the question, and the work is one multiply or one divide, with powers of ten riding along.
Here is Caspian's sample again: it loses 2 × 10⁻¹² kg of mass, all converted to energy.
The givens are always the same shape: a mass or an energy, the equation E = mc², and c = 3.00 × 10⁸ m/s.
Square c first: (3.00 × 10⁸)² = 9.00 × 10¹⁶. Keep that factor on paper.
For the sample: E = (2 × 10⁻¹²)(3.00 × 10⁸)² = 1.8 × 10⁵ J.
The number parts gave 2 × 9.00 = 18, and the powers of ten made 10⁴; 18 × 10⁴ J is written 1.8 × 10⁵ J.
To find m from E, substitute the known E into E = mc², then divide both sides by c².
Check the size: dividing an everyday energy by 9.00 × 10¹⁶ gives a very small mass, and multiplying a small mass by 9.00 × 10¹⁶ gives a very large energy. An answer that breaks this pattern signals a powers-of-ten slip.
Units ride the final line, always: joules for E, kilograms for m.
A fuel pellet loses 6.0 × 10⁻¹³ kg of mass, all converted to energy. Use E = mc², with c = 3.00 × 10⁸ m/s, to calculate the energy released.
Write down the values:
m = 6.0 × 10⁻¹³ kg, c = 3.00 × 10⁸ m/s, E = ?
Write down the equation:
E = mc²
Substitute in the values, and calculate:
E = (6.0 × 10⁻¹³) × (3.00 × 10⁸)². Squaring c gives 9.00 × 10¹⁶. Then 6.0 × 9.00 = 54, and the powers of ten make 10³. So E = 54 × 10³ J, written 5.4 × 10⁴ J.
The energy released is 5.4 × 10⁴ J.
A reaction releases 3.6 × 10⁶ J of energy, every joule from converted mass. Use E = mc², with c = 3.00 × 10⁸ m/s, to calculate the mass converted.
Write down the values:
E = 3.6 × 10⁶ J, c = 3.00 × 10⁸ m/s, m = ?
Write down the equation:
E = mc²
Substitute in the values, and solve for m:
3.6 × 10⁶ = m × 9.00 × 10¹⁶, so m = (3.6 × 10⁶) ÷ (9.00 × 10¹⁶). Then 3.6 ÷ 9.00 = 0.40, and the powers of ten make 10⁻¹⁰. So m = 0.40 × 10⁻¹⁰ kg, written 4.0 × 10⁻¹¹ kg.
The mass converted is 4.0 × 10⁻¹¹ kg.
A sample loses 4.0 × 10⁻¹² kg of mass, all converted to energy. Using E = mc² with c = 3.00 × 10⁸ m/s, calculate the energy released, in joules, to two significant figures.
A reaction releases 2.7 × 10⁴ J of energy from converted mass. Using E = mc² with c = 3.00 × 10⁸ m/s, calculate the mass converted, in kilograms, to two significant figures.
An experiment converts 5.0 × 10⁻¹⁰ kg of mass entirely to energy. Using E = mc² with c = 3.00 × 10⁸ m/s, calculate the energy released, in joules, to two significant figures.
A process releases 6.3 × 10⁵ J, every joule from converted mass. Using E = mc² with c = 3.00 × 10⁸ m/s, calculate the mass converted, in kilograms, to two significant figures.
An instrument records a crystal losing 8.0 × 10⁻¹⁴ kg of mass, all converted to energy. Using E = mc² with c = 3.00 × 10⁸ m/s, calculate the energy released, in joules, to two significant figures.
A storage site's books show 1.08 × 10¹⁴ J released, all of it from converted mass. Using E = mc² with c = 3.00 × 10⁸ m/s, calculate the mass converted, in kilograms, to two significant figures.
During a test, an instrument records a pellet losing 3.0 × 10⁻¹¹ kg of mass, all converted to energy. Using E = mc² with c = 3.00 × 10⁸ m/s, calculate the energy released, in joules, to two significant figures.
A process releases 8.1 × 10⁵ J of energy, every joule from converted mass. Using E = mc² with c = 3.00 × 10⁸ m/s, calculate the mass converted, in kilograms, to two significant figures.
A sealed capsule's mass falls by 6.0 × 10⁻¹² kg as it radiates energy away. Using E = mc² with c = 3.00 × 10⁸ m/s, calculate the energy radiated, in joules, to two significant figures.