How long would it take for a block of ice to melt in the sun?

[RajdeepSingh7]

I was wondering if it would be possible to get help with this question:

A 400 gram block of ice at -6°C is allowed to warm to room temperature which is 28°C.

If the sun provides 600 Joules of energy per square meter per second and all this energy was used to warm the ice block, calculate how long it would take to warm the ice block to 28°C.
If there was one square meter of surface exposed to the sun for this purpose.

Any contributions will be appreciated

Thanks

 

[Hootenanny]

Welcome to PF RajdeepSingh7,

Before you can receive assistance, you must show us what you have attempted thus far.

 

[RajdeepSingh7]

Would we have to find the total energy required by using the Specific Heat Capacity?

And then calculate the Time?

 

[Hootenanny]

Would we have to find the total energy required by using the Specific Heat Capacity?

And then calculate the Time?

Assuming that this process occurs at atmospheric pressure, one would also have to account for the latent heat.

 

[RajdeepSingh7]

Answer+Check

A 400 gram block of ice at -6°C is allowed to warm to room temperature which is 28°C.

Calculate the net amount of energy that had to flow into the block for this to happen. The following data will assist your calculation:

Specific heat capacity of liquid water = 4200 J Kg K
Latent Heat of fusion of ice : 3.34 x 10^5 JKg
Specific heat capacity of Ice = 2.10 x 10^3 JKg K


If the sun provides 600 Joules of energy per square meter per second and all this energy was used to warm the ice block, calculate how long it would take to warm the ice block to 28°C.
If there was one square meter of surface exposed to the sun for this purpose.



1)

Warming of Ice from -6C to 0C, Specific Heat:
Change Temperature = 6C
c=2.10 x 10^3
m = 0.4
Energy Transferred ( Q) = mc Change in T
Energy Transferred ( Q) = 0.4 x 2.10 x 10^3 x 6
Energy Transferred ( Q) = 5040 J


2)

Melting the Ice from 0C to 0C, Latent Heat of Fusion:

Lf=3.34 x 10^5
m = 0.4
Energy Transferred ( Q) = m x Lf
Energy Transferred ( Q) = 0.4 x 3.34 x 10^5
Energy Transferred ( Q) = 133600 J

3)

Warming of Water from 0C to 28C, Specific Heat:
Change Temperature = 28C
c=4200
m = 0.4
Energy Transferred ( Q) = mc Change in T
Energy Transferred ( Q) = 0.4 x 4200 x 28
Energy Transferred ( Q) = 47040J


4)

Total Energy Required : 5040 J + 133600 J + 47040J = 185680 J


Part B ))

185680 J divided by 600 J

185680 / 600 = 309.437 Seconds

309.437 Seconds / 60 = 5.158 Minutes

Please Check My Answer and My Working Out!