Find the mass and side length of unknown ice cube?

[Adam_9333]

Homework Statement

transfer 650 EJ (10^18 joules) into an ice cube of unknown size. the temperature of h20 increases from -10c to 20c.

Given:
1) q=mc(t2-t1)
2) for phase change q=m(F)

specific heat of ice (c)= 2.22x10^3 J.kg^-1.K-1
heat of fusion (f)= 3.33x10^5 J/kg
specific heat of water (c) = 4186 J.kg^-1.K^-1)

Homework Equations

what is the side length and mass of cube?

The Attempt at a Solution

First I would use eqn 1 for the heating of ice from (-10C to 0C): q=650EJ, c t= 10C and solve for mass?
then use second equation for phase change solid to liquid (0C) using q=650,f and also solve for mass?
use eqn 1 again and solve for m of water going from (0C to 20C) q=650EJ, c, t=20c

Then I would sum the masses and solve for volume?

 

[TSny]

It's good that you see to break the problem up into 3 parts. However, in your solution you have assumed that 650 EJ are added for each part. But, 650 EJ is the total amount of heat added for the entire process of going from -10 C to 20 C.

 

[Adam_9333]

I am not given any mass information so I cannot find Q of each step. any way to combine the equations? like Q= (m)(c)(t) + (m)(f)

 

[Chestermiller]

How much heat does it take to start out with 1 kg of ice at -10 C and end up with liquid water at 20 C?

 

[Adam_9333]

Thank you for the responses, I appreciate it! so I figure out how much energy it takes for 1kg of ice to go from -10c to 20c and then I can use that relationship to figure out mass with 650EJ?

 

[Chestermiller]

Thank you for the responses, I appreciate it! so I figure out how much energy it takes for 1kg of ice to go from -10c to 20c and then I can use that relationship to figure out mass with 650EJ?

Sure

 

[SteamKing]

650 EJ is the amount of energy used by the entire Newnited States in about 7 years. That's going to be some ice cube.

https://en.wikipedia.org/wiki/Joule#Exajoule