Question

Substance Specific Heat (J kg-1 K-1) Aluminum 900 Copper 390 Brass 380 Silver 230 Platinum 130 If 130 J of energy were added to 1 kg of each of these samples, which would experience the GREATEST temperature increase? A) Aluminum B) Brass C) Copper D) Platinum

Answer 1

D) Platinum

Step-by-step explanation:

I agree

Answer 2

Choice D) Platinum.

Step-by-step explanation:

The specific heat of a substance gives the amount of energy it takes to raise the temperature of

• unit mass (1 kilogram in this case) of the substance,
• by one unit (1 degree Kelvin in this case).

After energy of size
`Q` is added to a sample with specific heat
`c` and mass
`m`, the temperature change would be:

`\Delta T = \displaystyle (Q)/(c \cdot m)`.

For each of these samples,
`Q = 130\; \rm J`,
`m = 1\; \rm kg`. Apply this equation to find the change in their temperatures.

Aluminum

`c = \rm 900\; \rm J \cdot kg^(-1)\cdot K^(-1)`.

`\begin{aligned}\Delta T &= (Q)/(c \cdot m) \\ &= (130\; \rm J)/(900\; \rm J \cdot kg^(-1) \cdot K^(-1)) \approx 0.14\; \rm K\end{aligned}`.

Copper

`c = 390\; \rm J \cdot kg^(-1)\cdot K^(-1)`.

`\begin{aligned}\Delta T &= (Q)/(c \cdot m) \\ &= (130\; \rm J)/(390\; \rm J \cdot kg^(-1) \cdot K^(-1)) \approx 0.33\; \rm K\end{aligned}`.

Brass

`c = 380\; \rm J \cdot kg^(-1)\cdot K^(-1)`.

`\begin{aligned}\Delta T &= (Q)/(c \cdot m) \\ &= (130\; \rm J)/(380\; \rm J \cdot kg^(-1) \cdot K^(-1)) \approx 0.34\; \rm K\end{aligned}`.

Silver

`c = 230\; \rm J \cdot kg^(-1)\cdot K^(-1)`.

`\begin{aligned}\Delta T &= (Q)/(c \cdot m) \\ &= (130\; \rm J)/(230\; \rm J \cdot kg^(-1) \cdot K^(-1)) \approx 0.57\; \rm K\end{aligned}`.

Platinum

`c = 130\; \rm J \cdot kg^(-1)\cdot K^(-1)`.

`\begin{aligned}\Delta T &= (Q)/(c \cdot m) \\ &= (130\; \rm J)/(130\; \rm J \cdot kg^(-1) \cdot K^(-1)) \approx 1.0\; \rm K\end{aligned}`.

Hence, platinum would experience the greatest temperature increase.