The temperature of a 80.0 g sample of mate- rial increases from 15.0◦C to 42.0◦C when it absorbs 4.00 kJ of energy as heat. What is the specific heat of this material?
Answer in units of J/g · K.
q = mass x specific heat x (Tfinal-Tinitial).
To find the specific heat of the material, we can use the formula:
q = m * c * ΔT
Where:
q: amount of heat absorbed or released by the material (in joules, J)
m: mass of the material (in grams, g)
c: specific heat of the material (in J/g · K)
ΔT: change in temperature (in degrees Celsius, °C)
In this case, we know the values of q, m, and ΔT. We need to solve for c.
Given:
q = 4.00 kJ (converted to joules) = 4000 J
m = 80.0 g
ΔT = 42.0°C - 15.0°C = 27.0°C
Plugging in the values into the formula:
4000 J = 80.0 g * c * 27.0°C
Now, we need to isolate the specific heat (c) on one side of the equation. Divide both sides of the equation by (80.0 g * 27.0°C):
4000 J / (80.0 g * 27.0°C) = c
The resulting value will give us the specific heat of the material in J/g · K. Let's calculate it.
c = 1.85 J/g · K
Therefore, the specific heat of the material is 1.85 J/g · K.