Question

A 50kg block of nickel at 90°C is dropped into an insulated tank that contains 0.5 m3 of liquid water at 25°C. Determine the following. a)-Is this a control volume or closed system? b)- What is the specific heat of nickel? c)-What is the specific heat of liquid water? d)-What is the temperature when thermal equilibrium has been reached?

Answer 1

a) Control Volume

b) 0.44 kJ/kg/K

c) 4.18 kJ/kg/K

d) 32°C

Step-by-step explanation:

a. The system is a control volume because mass can be added and removed from the system.

b) Nickel has a heat capacity of 0.44 kJ/kgK

c) The heat capacity of water at standard conditions is 4.18 kJ/kgK

d)

The thermal equilibrium means that the system heat transfer of the nickel and the water are the same:

`Q_(water)=Q_(nickel)`

`m_(nickel)\cdot{c_(nickel)}\cdot{c_(nickel)}\cdot{T_(2)-T_(nickel)}=m_(water)\cdot{c_(water)}\cdot{T_(2)-T_(water)}`

`m_(nickel)\cdot{c_(nickel)}\cdot{c_(nickel)}\cdot{T_(2)-T_(nickel)}=density_(water)\cdot{volume_(water)}\cdot{c_(water)}\cdot{c_(water)}\cdot{T_(2)-T_(water)}`

`50\cdot{0.44}\cdot{(T_2-363)}=0.5\cdot{1000}\cdot{4.18}\cdot{(T_2-298)}`

`T_2=305.03`

The temperature will be 305.03 K which is 32°C

Answer 2

a).Control volume

b).
`C_(nickel)` = 502.416 J/kg-K

c).
`C_(water)` = 4.187 kJ/kg-K

d).
`T_(2)` = 87.91°C

Step-by-step explanation:

a). It is a control volume system because mass is varying in the system.

b). Specific heat of nickel is
`C_(nickel)` = 502.416 J/kg-K

c). Specific heat of water is
`C_(water)` = 4.187 kJ/kg-K

d).We know that

net energy transfer = change in internal energy

`m_(nickel)* c_(nickel)(T_(2)-T_(nickel))=m_(water)* c_(water)(T_(2)-T_(water))`

`m_(nickel)* c_(nickel)(T_(2)-T_(nickel))=(volume_(water)* density_(water))* c_(water)(T_(2)-T_(water))`

`50* 502.416* (T_(2)-90)=(0.5* 1000)* 4.187* (T_(2)-25)`

`25120.8* (T_(2)-90)=2093.5*  (T_(2)-25)`

`T_(2)` = 87.91°C