Help with physics question please

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In summary, Judy must add 0.225 kg of ice at -12.0 C to the thermos bottle to reach a equilibrium temperature of 75.0 C for the boiling water. The original mass of the steam at 100 C is 0.0306 kg. The final temperature will be 54.8 C when 1 kg of water at 20C is heated by the condensation of 30 g steam at 100 C.
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1.) Judy places .150 kg of boiling water in a thermos bottle. How many kilograms of ice at -12.0 C must Judy add to the thermos so that the equilbrium temperature of the water is 75.0 C?

2.) A 0.03 kg ice cube at 0C is placed in an insulated box that contains a fixed quantity of steam at 100 C. When thermal equilibrium of this closed system is established, its temperature is found to be 23C. Determine the original mass of the steam at 100 C.

3.) What will the final temperature if 1 kg of water at 20C is heated by the condensation of 30 g steam at 100 C?
 
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1.) Judy must add 0.225 kg of ice at -12.0 C to the thermos bottle.2.) The original mass of the steam at 100 C is 0.0306 kg.3.) The final temperature will be 54.8 C.
 
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I would be happy to assist with these physics questions. Let's start with the first one. To determine the amount of ice needed to reach an equilibrium temperature of 75.0 C, we can use the formula Q = mcΔT, where Q is the heat exchanged, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. In this case, the heat exchanged by the boiling water is equal to the heat absorbed by the ice. So, we can set up the equation as follows:

Qwater = Qice

mwatercwaterΔTwater = miceciceΔTice

Since the water is boiling, it will have a temperature of 100 C, and we want the final temperature to be 75.0 C. So, ΔTwater = 100 C - 75.0 C = 25.0 C. The specific heat capacity of water is 4.18 J/gC, and the specific heat capacity of ice is 2.09 J/gC. Therefore, we can solve for the mass of ice needed as follows:

mice = mwatercwaterΔTwater / (ciceΔTice)

= (0.150 kg)(4.18 J/gC)(25.0 C) / (2.09 J/gC)(-12.0 C)

= 0.071 kg

Therefore, Judy will need to add 0.071 kg of ice at -12.0 C to the thermos to reach an equilibrium temperature of 75.0 C.

Moving on to the second question, we can use the same formula to determine the mass of steam in the insulated box. Since the system reaches thermal equilibrium, we can set up the equation as follows:

Qice = Qsteam

miceciceΔTice = msteamcsteamΔTsteam

Since the ice starts at 0 C and reaches a final temperature of 23 C, ΔTice = 23 C - 0 C = 23 C. The specific heat capacity of steam is 1.84 J/gC. Therefore, we can solve for the mass of steam as follows:

msteam = miceciceΔTice / (csteamΔTsteam)

= (0.03 kg)(2.09 J/gC)(23 C) / (1.84 J
 

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