A small amount (18.0 g) of an unknown metal is placed into a calorimeter with 180. g of water (c = 1.00 cal/gC). The water is initially at a temperature of 20 C and the metal is initially at 95 C. The calorimeter reaches a final temperature of 22.0 C. What is the specific heat (in cal/gC) of the unknown metal?
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q lost= q gained
The Specific Heat formula is:
c = ΔQ / (m × ΔT)
c: Specific Heat , in J/(kg.K)
ΔQ: Heat required for the temperature change, in J
ΔT: Temperature change, in K
m: Mass of the object, in kg
I agree with @taramgrant0543664 ..
She is awesome...
I just wanted to ask whether the equation would be 18*c*(99-22)=180*1*(22-20)
And for futher knowledge of what that material would be,
so a table is being uploaded about different specific heat.
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Well 99 should be 95 (simple mistake)
As well the 18.0g needs to be converted into kilograms so 0.018kg and the 180g would be 0.18kg
Specific heat capacity's units contain kg not g so that's why you would have to convert it
Yes Yes... you are right..... @taramgrant0543664 .