Question

What is the specific heat of a substance that absorbs 3.23 x 10^3 joules of heat when a sample of 1.0 x10^2 g of the substance increases in temperature from 11.0c to 48.0c

Answer 1

the specific heat of the substance is approximately 0.876 J/(g°C).

Step-by-step explanation:

Data:

Q = 3.23 x 10^3 J

m = 1.0 x 10^2 g

ΔT = (48.0°C - 11.0°C) = 37.0°C

Solution:

3.23 x 10^3 J = (1.0 x 10^2 g) * c * 37.0°C

c = (3.23 x 10^3 J) / [(1.0 x 10^2 g) * 37.0°C]

c = (3.23 x 10^3 J) / (1.0 x 10^2 g * 37.0°C)

c ≈ 0.876 J/(g°C) ans.