Answer:
1.70 J/(g °C)
Step-by-step explanation:
Specific heat, Cp, is a measure of an object's ability to handle energy (heat).
Specific heat has basic units of Energy per (mass*Temperature Change). Since there are many different measures of energy (Joules, calories, etc.), mass (kg, g, mole, etc.) and temperature (F, C, and K) There are many different specific heats for the same material. They are equivalent, bur simply expressed in different units.
To answer this question, we'll use J/(g °C), since those are the units used in the question.
A low specific heat means that a smaller amount of energy will result in a larger temperature change.
The expression we can use is Q = mcΔT, where
Q is the energy, in Joules (J)
m is the mass
c is the specific heat (in J/(g °C)), and
ΔT is the change in temperature as a result of the energy. ΔT is T2 - T1, where
T1 is the initial temperature and T2 is the final temperature.
The data:
m = 25.0g
T1 = 16.0 °C
T2 = 22.1 °C
Q = 259 J
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Q = mcΔT
Let's rearrange to solve for c, the specific heat:
Q = mcΔT
c = Q/(mΔT)
Enter the data:
c = Q/(mΔT)
c = (259J)/((25.0g)*(22.1C - 16.0C)
c = 1.698 or 1.70 to 3 sig figs
c = 1.70 J/(g °C)