What is the specific heat in J/g Celsius for a metal sample with a mass 95.6 g which absorbs 841 J of energy when it's temperature increases from 30.0 degrees Celsius to 98.0 degrees Celsius?

To find the specific heat, we can use the formula:

q = m * c * ΔT

Where:
q is the heat energy absorbed or released by the substance,
m is the mass of the substance,
c is the specific heat of the substance,
ΔT is the change in temperature.

In this case, we know:
q = 841 J,
m = 95.6 g,
ΔT = (98.0 - 30.0) = 68.0 °C.

Now, we can rearrange the equation to solve for c:

c = q / (m * ΔT)

Plugging in the values, we have:

c = 841 J / (95.6 g * 68.0 °C)

To get the answer in J/g °C, we need to convert grams to kilograms:

c = 841 J / (0.0956 kg * 68.0 °C)

Now we can calculate the specific heat:

c = 841 J / (0.0956 kg * 68.0 °C)
c ≈ 1.31 J/g °C

Therefore, the specific heat for the metal sample is approximately 1.31 J/g °C.