9. A piece of iron (mass 25.0 g) at 398 K is placed in a styrofoam coffee cup containing 25.0 mL of water at 298 K. Assuming that no heat is lost to the cup or the surroundings, what will the final temperature of the water be? The specific heat capacity of iron 0.449 J/g°C and water = 4.18 J/g°C. A) 348 K B) 308 K C) 287 K Heat cap frec. het 9=CAT, C- śmi mass D) 325 K .0, v-4495/3°C [Tf - 398k] = - 25,09*4.18 Jlg°c [Tf-298k] E) 388 K Answer: B 5 ma 25.07 0.443T¢ - 173.702= -4.187F+

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Chapter7: Chemical Energy
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Problem 109AE: A sample of nickel is heated to 99.8C and placed in a coffee-cup calorimeter containing 150.0 g...
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In the following question, I used the q=ms(delta T) equation, and q_system=-q_surroundings, with the final temperature of water being the answer I'm solving for. With my current arithmetic I get an answer that's way off. And I think I may have my equation incorrectcly set up somehow, because when I plug in the correct answer to try working backwards, the q_system does not equal the -q_surroundings.

I'm not sure why my equation would be wrong though. Isn't the final temperature the same for both sides of the equation? 

9. A piece of iron (mass = 25.0 g) at 398 K is placed in a styrofoam coffee cup containing 25.0 mL of
water at 298 K. Assuming that no heat is lost to the cup or the surroundings, what will the final
temperature of the water be? The specific heat capacity of iron 0.449 J/g°C and water = 4.18 J/g°C.
A) 348 K
B) 308 K
C) 287 K
Heat cap. free. het
9=CAT, C-smemars
D) 325 K S.0, v 04495/9°C [Tf-39ak] = - 25,09*4.18 Jlg°c [Tf-298k]
E) 388 K
Answer: B
, m- 25.09
0-443Tif - 178.702= -4.187f+
25.0m
Transcribed Image Text:9. A piece of iron (mass = 25.0 g) at 398 K is placed in a styrofoam coffee cup containing 25.0 mL of water at 298 K. Assuming that no heat is lost to the cup or the surroundings, what will the final temperature of the water be? The specific heat capacity of iron 0.449 J/g°C and water = 4.18 J/g°C. A) 348 K B) 308 K C) 287 K Heat cap. free. het 9=CAT, C-smemars D) 325 K S.0, v 04495/9°C [Tf-39ak] = - 25,09*4.18 Jlg°c [Tf-298k] E) 388 K Answer: B , m- 25.09 0-443Tif - 178.702= -4.187f+ 25.0m
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