A 116-g cube of ice at 0 ∘∘C is dropped into 1.26 kg of water that...

A 116-g cube of ice at 0 ∘∘C is dropped into 1.26 kg of water that was originally 84.1∘C. What is the final temperature of the water after the ice melts and the water comes to thermal equilibrium? This problem requires a lot of algebra. You will make fewer errors if you solve for the answer using symbols and then plug in the numbers. The specific heat of water and the latent heat of fusion for water are given in tables in your reading assignment.

Expert Solution

As you know the loss of heat is equal to the gain of the heat.

ML +Ms(T-T1) = ms(T2-T) ..1

Here L =334000J/kg , s=4186 J/kg°C M=0.116 kg , m=1.26 kg , T1 =0°C , T2 =84.1°C , T is final temperature.

Put these values in the equation 1

0.116×334000+0.116×4186×(T-0) =1.26×4186×(84.1-T)

38744+485.576T =443,573.676-5274.36T

485.576T+5274.36T =443573.676-38744

5759.936T=404829.676

T=404829.676/5759.936

T=70.28°C. Answer

genius_generous answered 2 years ago

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