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QUESTION 19 1.3 kg of ice at 0 deg C is dropped into a cooler containing 14 kg of water at 1 7 deg C. Given a latent heat of fusion of water/ice of 79.7 cal/g and a specific heat of water of 1.00 cal/g deg C, what is the final temperature of the mixture (in deg。? QUESTION 20 In a similar situation to the previous problem, what is the smallest mass of ice (in kg) at 0 deg C that would have to be added to a cooler with 11.7 kg of water at 12 deg C to reduce the final temperature to 0 deg C?

Answer 1

19)

As you Know

Loss of the energy=gain of the energy.

dQ1=dQ2

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

Here M=1.3kg=1300g ,L=79.7cal/g, s=1cal/g℃,m=14000g

T1=0℃, T2=17℃, T = final temperature

Put these values in equation ...1

1300×79.7+1300×1×(T-0)=14000×1(17-T)

103610+1300T=238000-14000T

1300T+14000T=238000-103610

15300T=134390

T=134390/15300

T=​​8.784℃ Answer.

20)

Similarly

ML=ms(T1-T)

Here L=79.7cal/g, s =1 cal /g℃ ,m=11.7kg, T1=12℃, T=0℃

Put these values in above equation

79.7M=11.7×1×12

M=140.4/79.7

M=1.672 kg Answer.