Thermal Conductivity and latent heat

[risingabove]

Homework Statement

If a copper kettle has a base of thickness 2.0mm and an area 3.0 x 10^-2 m² estimate the steady difference in temperature between the inner and outer surface of the base which
must be maintained to enable enough heat too pass through so that the temperature of 1 kg of water
rises of 0.25 K/s assume that there are no heat losses.

(ii) After reaching the temperature of 373K the water is allowed to boil under the same conditions for 120 seconds and the mass of water remaining in the kettle is 0.948kg.
Deduce a value for the specific latent heat of vaporization of water ( neglecting condensation of the steam in the kettle)

Homework Equations

Thermal conductivity of copper 3.8 x 10² W/m/C
Specific Heat Capacity of water 4.0 x 10³ J/kg/K

Equations used

dQ/dt = kA(dΘ/dx)

where dQ/dt = rate of flow of heat
K = Thermal Conductivity
A = Area
dΘ = Change in Temperature
dx = Thickness
whereby dΘ/dx = Temperature gradient

Q = mCT

Where Q = Heat require
m = Mass of substance
C = Specific heat capacity
T = Temperature

Q= mL

Where Q = Heat Require
m = Mass
L = Latent Heat

The Attempt at a Solution

Firstly i used the specific heat capacity equation to find the rate of flow of heat

Q = 1 (4.2 x 10³ J/kg/K ) 0.25 K/s
= 1050 J/s

Using the above answer i substitute it into the equation for thermal conductivity

1050 = 3.8 x 10² W/m/C (3.0 x 10^-2 m² )(dΘ/0.002m)
by simple calculations my answer = 0.18 degrees

Part (ii)

I am abit lost as to where to go from here... but I am thinking i will have to multiply 120 seconds by 0.25K/s which = 30 K/s

and used the 30K/s in the specific heat capacity equation ...( but I am thinking that doesn't make any sense)

By using the 373k and substituting it into the heat capacity equation

Q = 1(4.2 x 10³ J/kg/K)373
= 1566600J

and using the answer above in the Latent heat of vaporization equation

1566600 = 0.948L
L = 1.6 x 10⁶ J/kg/K

This seems wrong tho...

By the way is my answer correct for Part (i) ??

 

[DrClaude]

I am abit lost as to where to go from here... but I am thinking i will have to multiply 120 seconds by 0.25K/s which = 30 K/s

The temperature will be constant, so it's abit strange to start with something in K/s!

In part (i), you figured out how much power is furnished to the kettle. Start from there.

By the way is my answer correct for Part (i) ??

I didn't check all the calculations, but the approach seemed right.

 

[risingabove]

YAYYYYY i got!...geezeeee that was simple enough...

ok this is what i did ...please let me kno if I am correct...although i think i am

soo you said to start from the power i got in part (i) which was 1050 J/s

Therefore for rate of flow of heat( which is power) dQ/dt = 1050watts

since the time was given in the problem, 120seconds, i substituted it in for dt

solving for dQ i got 12600

then by subtracting the initial mass from the final, i got the mass of water that was vaporized
= 0.052

these values was then substituted into the formula for latent heat of vaporization

giving me 2.4 x 10⁶

 

[DrClaude]

these values was then substituted into the formula for latent heat of vaporization

giving me 2.4 x 10⁶

Great, but don't forget the units