Calculating Specific Heat: 13g Sample Loses 25.8J

[Jam22]

When 13 grams of a sample cools from 56◦C to 34.4◦C it loses 25.8 Joules of heat. What is the specific heat of the sample? Answer in units of J/g ·◦ C.

I know it seems simple, but I completely forgot how to do this. Please help.

 

[Subdot]

The specific heat formula is q = mc$\Delta$T, where q is the amount of heat absorbed or released by an object, m is the object's mass, c is the object's specific heat, and $\Delta$T is the change in temperature that the object experiences (in other words, the final temperature minus the initial temperature of the object).

Simply plug in your values and solve for c to get the specific heat. Just note that when you are given the magnitude of the heat of the object (such as in this problem), you must determine whether the heat you were given takes a positive or negative value in that equation, then plug in that value for q.

 

[Blueshift5]

To calculate the specific heat of the sample, we can use the formula Q = mcΔT, where Q is the amount of heat lost, m is the mass of the sample, c is the specific heat, and ΔT is the change in temperature.

In this case, we have Q = -25.8J, m = 13g, ΔT = (34.4-56)◦C = -21.6◦C.

Substituting these values into the formula, we get:

-25.8J = (13g) c (-21.6◦C)

Solving for c, we get:

c = -25.8J / (13g * -21.6◦C)

c = 0.116 J/g ·◦ C

Therefore, the specific heat of the sample is 0.116 J/g ·◦ C. This means that it takes 0.116 Joules of energy to raise the temperature of 1 gram of the sample by 1 degree Celsius.