Calculate the specific heat capacity of this metal

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SUMMARY

The discussion centers on calculating the specific heat capacity of a metal using the formula Q = mc(ΔT). A 35-gram piece of metal at 100°C loses 775 calories of heat while cooling to 30°C, leading to a temperature change (ΔT) of 70°C. The correct calculation yields a specific heat capacity of approximately 0.316 cal/g°C, as confirmed by multiple attempts and corrections in the forum. Users emphasized the importance of accurately applying the formula to avoid errors in the final answer.

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  • Understanding of the specific heat capacity formula Q = mc(ΔT)
  • Basic knowledge of calorimetry principles
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  • Familiarity with temperature scales (Celsius)
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shann0nsHERE
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When a 35 gram piece of some metal at 100°C is placed in water, it loses 775 calories of heat while cooling to 30°C. Calculate the specific heat capacity of this metal.

Q = mc(DELTA)T

T = 70 degrees C
m = 35 g
Q = 775 cal ??


When i plugged it in i got 0.34 but the program said it was wrong...and then i tried


c- 775 = (70)(35)c
c = 2.90

but I'm afraid to enter this answer because I only have one more chance to get it right...
 
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shann0nsHERE said:
When a 35 gram piece of some metal at 100°C is placed in water, it loses 775 calories of heat while cooling to 30°C. Calculate the specific heat capacity of this metal.

Q = mc(DELTA)T

T = 70 degrees C
m = 35 g
Q = 775 cal ??


When i plugged it in i got 0.34 but the program said it was wrong...and then i tried


c- 775 = (70)(35)c
c = 2.90

but I'm afraid to enter this answer because I only have one more chance to get it right...

When i plugged everything in i got c=.316 = 775/(35*70)
 
Thank you so so so much!
 

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