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.