Calculate the specific heat capacity of this metal

AI Thread Summary
To calculate the specific heat capacity of a metal, the formula Q = mc(ΔT) is used, where Q is the heat lost, m is the mass, and ΔT is the change in temperature. In this case, a 35 gram piece of metal cools from 100°C to 30°C, losing 775 calories of heat. The temperature change (ΔT) is 70°C, leading to the equation 775 = (35)(c)(70). After various calculations, the correct specific heat capacity was determined to be approximately 0.316 cal/g°C. The discussion highlights the importance of careful calculation and verification when solving for specific heat capacity.
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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