q = (m)(Dt)(cp) where q is the energy in joules

m is the mass

Dt is the change in initial temperature to the final temperature

cp is the specific heat (heat required to raise 1 g of the substance 1C)

q = (m)(Dt)(cp)

2.5 x 10

[1.0 x 10

2.5 x 10

^{3}J = (1.0 x 10^{4}g)(70C-10C)(cp)__[2.5 x 10__= cp^{3}J][1.0 x 10

^{4}g][70C-10C] cp=

gC

__0.0042 J__gC

*Answer: cp = 0.0042 J/gC*2) A 1.0 kg sample of metal with a specific heat of 0.50 KJ/KgC is heated to 100.0C and then placed in a 50.0 g sample of water at 20.0C. What is the final temperature of the metal and the water?

q

_{metal}= (m)(Dt)(cp) q_{water}= (m)(Dt)(cp) q

_{metal = }q_{water}_{ }(m)(Dt)(cp) = (m)(Dt)(cp)

__[1.0Kg][100C-T__=

_{f}][0.50KJ]__[50.0g][T__

_{f}-20C][4.184 J][ KgC] [ gC ]

__[1.0Kg][1000g][100C-T__=

_{f}][0.50KJ][ 1Kg][1000J]__[50.0g][T__

_{f}-20C][4.184 J][ 1Kg] [ KgC][1000g][ 1KJ] [ gC ]

__[1000g][100C-T__=

_{f}][0.50J]__[50.0g][T__

_{f}-20C][4.184 J][ gC] [ gC ]

[

__1000g][100C-T__=_{f}][0.50 J/gC]__[__T_{f}-20C__]__

[ 50.0g] [4.184 J/gC] [2.39][100C-T

_{f}] =__[__T_{f}-20C__]__ 239C-2.39T

_{f}= T_{f}-20C 239C + 20C = T

_{f }+ 2.39T_{f } 259C = 3.39T

_{f}__259C__= T

_{f }3.39

76C = T

_{f }

*Answer: Final temperature of metal and water is 76C* q

_{metal}= (m)(Dt)(cp) q_{water}= (m)(Dt)(cp) q

_{metal = }q_{water}_{ }(m)(Dt)(cp) = (m)(Dt)(cp)

__[2.80Kg][100C-T__=

_{f}][0.43KJ]__[50.0g][T__

_{f}-30C][4.184 J][ KgC] [ gC ]

__[2.8.0Kg][1000g][100C-T__=

_{f}][0.43KJ][ 1Kg][1000J]__[50.0g][T__

_{f}-30C][4.184 J][ 1Kg] [ KgC][1000g][ 1KJ] [ gC ]

__[2800g][100C-T__=

_{f}][0.43J]__[50.0g][T__

_{f}-30C][4.184 J][ gC] [ gC ]

[

__2800g][100C-T__=_{f}][0.43 J/gC]__[__T_{f}-30C__]__

[ 50.0g] [4.184 J/gC] [24.08][100C-T

_{f}] =__[__T_{f}-30C__]__ 2408C-24.08T

_{f}= T_{f}-30C 2408C + 30C = T

_{f }+ 24.08T_{f } 2438C = 25.08T

_{f}__2438C__= T

_{f }325.08

97C = T

_{f }

*Answer:*

*final temperature of metal and water is 97C*
the solutions to problems can be simply understood.. :)

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