The heat capacity of lead is 0.13 J/g•°C. How many joules of heat would be required to raise the temperature of 15 g of lead from 22°C to 37°C?
q = mass Pb x specfic heat Pb x (Tfinal-Tinitial)
To calculate the amount of heat required to raise the temperature of a substance, we can use the formula:
Q = m * C * ΔT
Where:
Q is the amount of heat in joules (J)
m is the mass of the substance in grams (g)
C is the specific heat capacity of the substance in J/g•°C
ΔT is the change in temperature in °C
In this case, we have:
m = 15 g
C = 0.13 J/g•°C
ΔT = 37°C - 22°C = 15°C
Substituting these values into the formula, we can calculate the amount of heat:
Q = 15 g * 0.13 J/g•°C * 15°C
Q = 292.5 J
Therefore, 292.5 joules of heat would be required to raise the temperature of 15 g of lead from 22°C to 37°C.
To calculate the amount of heat required to raise the temperature of a substance, we can use the formula:
Q = m * C * ΔT
Where:
Q is the amount of heat transferred (in Joules),
m is the mass of the substance (in grams),
C is the specific heat capacity of the substance (in J/g•°C),
ΔT is the change in temperature (in °C).
In this case, we are given:
m = 15 g (mass of lead)
C = 0.13 J/g•°C (specific heat capacity of lead)
ΔT = 37°C - 22°C (change in temperature)
Substituting the values into the formula:
Q = 15 g * 0.13 J/g•°C * (37°C - 22°C)
Q = 15 g * 0.13 J/g•°C * 15°C
Q = 292.5 J
Therefore, it would require 292.5 Joules of heat to raise the temperature of 15 grams of lead from 22°C to 37°C.