How much energy is required to raise the temperature of 3 kg of lead from 15C to 20C? Use the table below and this equation: Q = mcT.
To calculate the energy required, we need to use the formula Q = mcΔT, where:
- Q is the energy required,
- m is the mass of the substance (3 kg),
- c is the specific heat capacity of the substance (given in the table below), and
- ΔT is the change in temperature (20°C - 15°C).
The specific heat capacity of lead is 130 J/kg°C.
Now we can substitute the values into the formula and solve for Q:
Q = (3 kg) * (130 J/kg°C) * (20°C - 15°C)
Q = 3 kg * 130 J/kg°C * 5°C
Q = 1950 J * 5
Q = 9750 J
Therefore, it would require 9750 Joules of energy to raise the temperature of 3 kg of lead from 15°C to 20°C.
To calculate the amount of energy required to raise the temperature of a substance, we can use the equation Q = mcΔT, where:
- Q is the amount of energy in joules,
- m is the mass of the substance in kilograms,
- c is the specific heat capacity of the substance in joules per kilogram degree Celsius,
- ΔT is the change in temperature in degrees Celsius.
In this case, we have 3 kg of lead, a change in temperature of 20°C - 15°C = 5°C, and we need to find the amount of energy (Q) required.
From the given table, we can find the specific heat capacity of lead. Assuming the specific heat capacity of lead is 130 J/kg°C, we can substitute the known values into the equation:
Q = (m)(c)(ΔT)
Q = (3 kg)(130 J/kg°C)(5 °C)
Simplifying the equation, we can solve for Q:
Q = (3)(130)(5) J
Q = 1950 J
Therefore, the amount of energy required to raise the temperature of 3 kg of lead from 15°C to 20°C is 1950 joules.