Question
What is the specific heat of a substance that absorbs 2.5 x 10 3joules of heat when a sample of 1.0 x 10 4g of the substance increases in temperature from 10.0°C to 70.0°C?
Answers
GPT 3.5
The formula to calculate the specific heat is:
q = m * c * ΔT
where:
q is the heat absorbed/released (in joules),
m is the mass of the substance (in grams),
c is the specific heat (in J/g°C), and
ΔT is the change in temperature (in °C).
Given:
q = 2.5 x 10^3 J
m = 1.0 x 10^4 g
ΔT = (70.0°C - 10.0°C) = 60.0°C
Rearranging the equation:
c = q / (m * ΔT)
Calculating the specific heat:
c = (2.5 x 10^3 J) / (1.0 x 10^4 g * 60.0°C)
c ≈ 4.17 J/g°C
Therefore, the specific heat of the substance is approximately 4.17 J/g°C.
q = m * c * ΔT
where:
q is the heat absorbed/released (in joules),
m is the mass of the substance (in grams),
c is the specific heat (in J/g°C), and
ΔT is the change in temperature (in °C).
Given:
q = 2.5 x 10^3 J
m = 1.0 x 10^4 g
ΔT = (70.0°C - 10.0°C) = 60.0°C
Rearranging the equation:
c = q / (m * ΔT)
Calculating the specific heat:
c = (2.5 x 10^3 J) / (1.0 x 10^4 g * 60.0°C)
c ≈ 4.17 J/g°C
Therefore, the specific heat of the substance is approximately 4.17 J/g°C.