How much heat must be removed from 350 g of 20.0 degree C liquid water to turn i

Question

How much heat must be removed from 350 g of 20.0 degree C liquid water to turn it into 0 degree C solid ice? The choices below are the amount of heat required. A. 87, 248 J B. 29, 302 J C. 69, 293 J D. 145, 852 J E. 116, 550 J F. 54, 873 J An insulated 80 g aluminum calorimeter cup contains 175 g of water. The cup and water are initially a 20.0 degree C. A "mystery" metal with a mass of 150 g and at a temperature of a 180 degree C is placed into this cup. After some time passes the three substances reach an equilibrium temperature of 31.4 degree C. What is the specific heat of this "mystery" metal? A. 864 J/(kg degree C) B. 411 J/(kg degree C) C. 658 J/(kg degree C) D. 489 J/(kg degree C) E. 382 J/(kg degree C) F. 548 J/(kg degree C)

Explanation / Answer

To convert water to ice

1. First we need to bring its temperature from 20oC to 0oC

Heat required for this process = H1= (specific heat of water) * (mass of water) * ( change in Temperature)

Specific heat of water = 4.2 J/g-oC

mass of water = 350 g

Change in temperature = 20 - 0 = 20 oC

so

H1  = 4.2* 350 * 20 = 29260 J

2. Now we need to remove Latent heat this heat is given by

H2 = Latent heat of fusion * mass of water

Latent heat of fusion = 333.5 J

Hence

H2 = 333.5 * 350 = 116725 J

So total heat required to be removed = 29260 + 116725 = 145985 J

Hence correct answer is D , error in answer calculated above is due to approximate values taken for latent heat of fusion and specific heat , which may differ in small amount as mentioned in your textbook.

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