Answer:
Choice D) Platinum.
Explanation:
The specific heat of a substance gives the amount of energy it takes to raise the temperature of
After energy of size [tex]Q[/tex] is added to a sample with specific heat [tex]c[/tex] and mass [tex]m[/tex], the temperature change would be:
[tex]\Delta T = \displaystyle \frac{Q}{c \cdot m}[/tex].
For each of these samples, [tex]Q = 130\; \rm J[/tex], [tex]m = 1\; \rm kg[/tex]. Apply this equation to find the change in their temperatures.
[tex]c = \rm 900\; \rm J \cdot kg^{-1}\cdot K^{-1}[/tex].
[tex]\begin{aligned}\Delta T &= \frac{Q}{c \cdot m} \\ &= \frac{130\; \rm J}{900\; \rm J \cdot kg^{-1} \cdot K^{-1}} \approx 0.14\; \rm K\end{aligned}[/tex].
[tex]c = 390\; \rm J \cdot kg^{-1}\cdot K^{-1}[/tex].
[tex]\begin{aligned}\Delta T &= \frac{Q}{c \cdot m} \\ &= \frac{130\; \rm J}{390\; \rm J \cdot kg^{-1} \cdot K^{-1}} \approx 0.33\; \rm K\end{aligned}[/tex].
[tex]c = 380\; \rm J \cdot kg^{-1}\cdot K^{-1}[/tex].
[tex]\begin{aligned}\Delta T &= \frac{Q}{c \cdot m} \\ &= \frac{130\; \rm J}{380\; \rm J \cdot kg^{-1} \cdot K^{-1}} \approx 0.34\; \rm K\end{aligned}[/tex].
[tex]c = 230\; \rm J \cdot kg^{-1}\cdot K^{-1}[/tex].
[tex]\begin{aligned}\Delta T &= \frac{Q}{c \cdot m} \\ &= \frac{130\; \rm J}{230\; \rm J \cdot kg^{-1} \cdot K^{-1}} \approx 0.57\; \rm K\end{aligned}[/tex].
[tex]c = 130\; \rm J \cdot kg^{-1}\cdot K^{-1}[/tex].
[tex]\begin{aligned}\Delta T &= \frac{Q}{c \cdot m} \\ &= \frac{130\; \rm J}{130\; \rm J \cdot kg^{-1} \cdot K^{-1}} \approx 1.0\; \rm K\end{aligned}[/tex].
Hence, platinum would experience the greatest temperature increase.
