How Much Ice is Required to Cool a Whale's Tank from 20°C to 10°C?

[mizzy]

Homework Statement

What mass of ice at -10.0 degrees celsius is needed to cool a whale's water tank, holding 1.20 x 10^3 m^3 of water, from 20.0 degrees down to a more comfortable 10 degrees.

Homework Equations

Q = mc *delta T

Q = mL

The Attempt at a Solution

i'm trying to fill out a table with knowns and unknowns before I actually solve the problem. This is what i got, can someone help me with the temperatures?

ICE: mass = x
c = 2090 J/kg C
Ti = -10.0
Tf = 0

MELT: mass = x
L = 3.33 x 10^5

ICE - WATER: mass = x
c = 4190
This is where I'm stuck what's the temperature?
Ti = ?
Tf = ?

WATER: mass = ??
c= 4190
Ti = 20.0
Tf = 10


Can someone let me know where i went wrong?? I still need to find the mass of water.

Thanks!

 

[iratern]

Do you know the concept of density? as in mass per volume? you can use that to calculate your waters mass.

As for the final temperature of the water you should calculate how much heat it needs to give up. Then see how much heat the ice can give up...

 

[mizzy]

Yes, i know about density. To get mass, just multiply density of water and volume given.

 

[iratern]

then you can use that to get the heat water needs to give for the temperature drop. Then you can use algebra to solve for the change in the ices temperture/ melting.

 

[rwisz]

Hi there, it looks like you're on the right track! The key to solving this problem is to use the heat equation, Q = mc * delta T, for each step of the process. Let's break down the problem step by step:

1. First, we need to determine the amount of heat (Q) that needs to be removed from the water tank in order to cool it from 20 degrees to 10 degrees. We can do this by using the mass of water (1.20 x 10^3 m^3) and the specific heat capacity of water (4190 J/kg C):

Q = (1.20 x 10^3 m^3) * (4190 J/kg C) * (20 C - 10 C)

= 5.028 x 10^7 J

2. Next, we need to find the amount of heat (Q) that can be removed by the ice as it melts. We can use the mass of ice (x), the latent heat of fusion (3.33 x 10^5 J/kg), and the change in temperature (from -10 degrees to 0 degrees) to calculate this:

Q = x * (3.33 x 10^5 J/kg) * (0 C - (-10 C))

= 3.33 x 10^6 x J

Note that we need to use the latent heat of fusion since the ice will be melting as it cools the water.

3. Now, we can set these two Q values equal to each other and solve for the mass of ice (x):

5.028 x 10^7 J = 3.33 x 10^6 x J

x = 15.1 kg

So, 15.1 kg of ice will be needed to cool the water tank from 20 degrees to 10 degrees.

4. Finally, we can use the remaining mass of water (1.20 x 10^3 m^3) and the specific heat capacity of water (4190 J/kg C) to calculate the mass of water that will be left after the ice melts:

Q = (1.20 x 10^3 m^3) * (4190 J/kg C) * (10 C - 0 C)

= 5.016 x 10^7 J

This is the amount of heat that will be removed by the remaining water.