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mtimko
 one year ago
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?
mtimko
 one year ago
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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taramgrant0543664
 one year ago
Best ResponseYou've already chosen the best response.3q lost= q gained The Specific Heat formula is: c = ΔQ / (m × ΔT) q=cmΔT Where: 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

taramgrant0543664
 one year ago
Best ResponseYou've already chosen the best response.3cmΔT=cmΔT lost= gained

arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.0I agree with @taramgrant0543664 .. She is awesome... I just wanted to ask whether the equation would be 18*c*(9922)=180*1*(2220) And for futher knowledge of what that material would be, so a table is being uploaded about different specific heat.

taramgrant0543664
 one year ago
Best ResponseYou've already chosen the best response.3Well 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

arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.0Yes Yes... you are right..... @taramgrant0543664 . Got it!
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