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1 year ago

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the specific heat of the metallic element rhodium is 0.240 j/g-k at 25°c. if 586 j of heat is added to an 80.0 g block of rhodiu

m metal that is initially at 25.0°c, what is the expected final temperature of the block? the specific heat of the metallic element rhodium is 0.240 j/g-k at 25°c. if 586 j of heat is added to an 80.0 g block of rhodium metal that is initially at 25.0°c, what is the expected final temperature of the block? 2470°c 30.5°c 55.5°c 26.8°c 32.3°c

1 answer:

andrey2020 [161]1 year ago

6 0

The expected final temperature of the block, given that 586 J of heat were added to it is 55.5 °C

<h3>How to determine the final temeprature</h3>

We'll begin by obtaining the change in the temperature of the block. This can be obtained as follow:

Specific heat capacity of block (C) = 0.240 J/gºC
Heat added (Q) = 586 J
Mass of block (M) = 80.0 g
Change in temperature (ΔT) =?

Q = MCΔT

Divide both sides by MC

ΔT = Q / MC

ΔT = 586 / (80.0 × 0.240)

ΔT = 586 / 19.2

ΔT = 30.5 °C

Finally, we shall determine the final temperature of the block. This can be obtained as follow:

Initial temperature (T₁) = 25 °C
Change in temperature (ΔT) = 30.5 °C
Final temperature (T₂) = ?

ΔT = T₂ – T₁

30.5 = T₂ – 25

Collect like terms

T₂ = 30.5 + 25

T₂ = 55.5 °C

Thus, from the calculation made above, we can conclude that the final temperature is 55.5 °C

Learn more about heat transfer:

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