What amount of heat will be removed if a body of heat capacity 380jk^-1 is cooled from 50°c to 20°c
To calculate the amount of heat removed, we need to use the formula:
Q = mcΔT
Where:
Q = amount of heat removed
m = mass of the body (assume 1 kg)
c = heat capacity of the body (380 J/k·kg)
ΔT = change in temperature (50°C - 20°C = 30°C)
Substituting the given values into the formula:
Q = (1 kg)(380 J/k·kg)(30°C)
Q = 11,400 J
Therefore, 11,400 Joules of heat will be removed.
To calculate the amount of heat removed, you can use the equation:
Q = mcΔT
Where:
Q = amount of heat removed
m = mass of the body
c = specific heat capacity of the body
ΔT = change in temperature
Since the question states that the body has a heat capacity of 380 J/k^-1 (Joules per Kelvin), we can assume that the mass is 1 kg (since heat capacity is defined as the amount of heat required to raise the temperature of 1 kg of the substance by 1 K).
ΔT = 50°C - 20°C = 30°C
Converting ΔT to Kelvin:
ΔT = 30 K
Substituting the values into the equation:
Q = 1 kg * 380 J/k^-1 * 30 K
Q = 11,400 J
Therefore, the amount of heat removed from the body is 11,400 Joules.