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December 18, 2014

December 18, 2014

Posted by **Sara** on Friday, November 17, 2006 at 7:54pm.

Look up the specific heat of aluminum. Multiply it by 0.20 kg and 165 C to clculate the amount of heat transferred TO the aluminum as it equilibrates at 0 C. Call that amount of heat Q.

Then write an equation for the amount of heat transferred FROM 3.0 C H2O as it freezes at 0 C. Set it equal to Q. The mass M of water will be the only unknown. Solve for it

specific heat of aluminum is 9x10^2

so this is what I did

(.26)(900)(165)=m(4186)(3)

specific heat of water (15C) is 4186

but the answer turned out to be incorrect

I tried it again with a diff. value of specific heat which is 2000

but it was still incorrect,

CAN SOME1 PLZ HELP ME OUT

We left out a step here. The water does cool from 3 to 0. Then some of it FREEZES. The heat of fusion of water to ice is 334 J/g. We need to calcualate this in. So, the difference in the amount of heat gained by the aluminum and the amount of heat lost by the water as it cools is the heat of fusion of whatever freezes!

It is important to show the units you are using. A specific heat of water of 4186 would correspond to units Joules per kg degree C. DanH's comments about the heat of fusion must also be taken into account. Please show your work for further assistance. The specific heat of aluminum is actually 897 J/kg C

I am still confused

Can some1 write it out for me?

what I did is

(897 J/kg C)(.26kg)(165C)=m(334 J/g)3C

Is this correct

????????

No, it isn't correct.

(897 J/kg*c)(0.26 kg)(165) + you omitted the (1.4 kg)(4186 J/kg*c)(0-3) = m(334,000 J/kg*c).

The 334 J/g*c must be changed to J/kg*c.

Solve for m

I tried the formula given above...but the answer is still wrong...how come?

never mind. i used the wrong number for something. thank you!

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