Question

A metal block of mass 7kg at 873k is dropped into30kg of liquid at 313k. if the heat capacity of metal and the liquid are 23.59j/k'g'and 5.02j/k/g and that no heat is lost to the surrounding, calculate the entropy of the metal block, the liquid and total entropy change

Answer 1

This is a thermo-equillibrium situation. We can use the equation

Loss of Heat of the Metal = Gain of Heat by the Water

−

Q

m

=

+

Q

w

Q

=

m

Δ

T

C

p

Q

=

Heat

m

=

mass

Δ

T

=

(

T

f

−

T

i

)

T

f

=

Final Temp

T

i

=

Initial Temp

C

P

=

Specific Heat

Metal

Water

m

=

17.5

g

T

f

=

30.0

o

C

T

i

=

125.0

o

C

C

P

=

x

Water

m

=

15.0

g

T

f

=

30.0

o

C

T

i

=

25.0

o

C

C

P

=

4.184

J

g

o

C

#

−

Q

m

=

+

Q

w

−

[

m

(

T

f

−

T

−

i

)

C

p

]

=

m

(

T

f

−

T

−

i

)

C

p

−

[

17.5

g

(

30

o

C

−

125

o

C

)

x

]

=

15

g

(

30

o

C

−

25

o

C

)

4.184

J

g

o

C

−

[

17.5

g

(

−

95

o

C

)

x

]

=

15

g

(

5

o

C

)

)

4.184

J

g

o

C

(

1662.5

g

o

C

)

x

=

313.8

J

1662.5

g

o

C

x

1662.5

g

o

C

=

313.8

J

1662.5

g

o

C

C

p

=

0.189

J

g

o

C