When a  brass black was dropped into a 124-gram water, the water temperature increased from 21.0°C to 36.8°C. What is the heat capacity of the brass block? The initial temperature of the brass was 314.2°C and the specific heat of water is 4.184J/gk. If the mass of the brass block is 77.764g, what is the specific heat of the brass?

Question
Asked Oct 29, 2019

When a  brass black was dropped into a 124-gram water, the water temperature increased from 21.0°C to 36.8°C. What is the heat capacity of the brass block? The initial temperature of the brass was 314.2°C and the specific heat of water is 4.184J/gk. If the mass of the brass block is 77.764g, what is the specific heat of the brass?

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Step 1

Given,

Mass of water, mwater = 124 g

Mass of brass, mbrass = 77.764 g

Initial temperature of water = 21.0°C = 294 K

Initial temperature of brass = 314.2°C = 587.2 K

Final temperature = 36.8°C = 309.8 K

Specific heat of water, cwater = 4.184 J/gK

The change in temperature for water and brass can be calculated as:

ATbrass Final temperature - Initial temperature of brass
ATbrass
309.8 K - 587.2 K = -277.4 K
ATwater Final temperature - Initial temperature of brass
ATwater= 309.8 K - 294 K = 15.8 K
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ATbrass Final temperature - Initial temperature of brass ATbrass 309.8 K - 587.2 K = -277.4 K ATwater Final temperature - Initial temperature of brass ATwater= 309.8 K - 294 K = 15.8 K

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Step 2

The heat lost by brass (qlost) when immersed in water will be equal to heat ...

qlost gain
= (mwater X Cwater ATwater
-(mbrass X Cbrass X ATbrass)
-[77.764 g
x Cbrass X (-277.4 K)] = 124 g x 4.184 J/gK x 15.8 K
x 15.8 K
gK
124 g x 4.184
Cbrass
77.764 g x 277.4 K
0.380 J/gK
Cbrass
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Image Transcriptionclose

qlost gain = (mwater X Cwater ATwater -(mbrass X Cbrass X ATbrass) -[77.764 g x Cbrass X (-277.4 K)] = 124 g x 4.184 J/gK x 15.8 K x 15.8 K gK 124 g x 4.184 Cbrass 77.764 g x 277.4 K 0.380 J/gK Cbrass

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