- #1
shann0nsHERE
- 6
- 0
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...
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...