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?

q = mc*delta T

To find the specific heat of the metal sample, we can use the formula:

q = mcΔT

where:
q is the heat absorbed by the substance (in this case, the metal sample)
m is the mass of the substance
c is the specific heat capacity of the substance
ΔT is the change in temperature

In this case, we are given:
q = 841 J
m = 95.6 g
ΔT = 98.0 °C - 30.0 °C = 68.0 °C

Now, let's rearrange the formula to solve for c:

c = q / (m * ΔT)

Substituting the given values:

c = 841 J / (95.6 g * 68.0 °C)

To get the specific heat in J/g Celsius, we divide the result by the mass (95.6 g):

c = (841 J / (95.6 g * 68.0 °C)) / 95.6 g

Now, we can cancel out the grams in the numerator and denominator:

c = 841 J / (95.6 * 68.0 °C)

Calculating the value:

c ≈ 0.132 J/g °C

Therefore, the specific heat of the metal sample is approximately 0.132 J/g °C.