540g of ice at 0°C is mixed with 540g of water at 80°C. The final temperature of the mixture is..???

2 Answers
Hriman
Feb 24, 2018

#tf=0.086°C#

Explanation:

#mL+mcΔt=-mcΔt#

#(540g)(334J/g)+(540g)*(4.184J/(g°C))(tf-0°)=-(540 g)(4.184 J/(g°C))(tf-80°)#

#180360J+2259.36 J/(°C)*tf-0J= -2259.36 J/(°C)+180748.8J#

#4518.72 J/(°C)*tf=388.8J#

#tf=0.086°C#

Jane
Feb 24, 2018

#0^@C#

Explanation:

Heat energy liberated by #540 g# of water to cool down to #0^@C# is #540*1*(80-0) C=43200 C# (using, #H = ms d theta# ,where, #m# is the mass, #s# is the specific heat and #d theta # is change in temperature)

Now,heat energy required for #540 g# of ice to get converted to same amount of water at #0^@C# is #540*80=43200 C# (using, #H=ml# where, #l# is the latent heat of melting of ice = #80 C/g#)

So,the required heat energy for the given amount of ice to get converted to water at #0^@C# will be supplied exactly due to the cooling of water to #0^@C#, so total #540+540=1080g# of water will be present at #0^@C#

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