SAT Subject Physics Practice Question 130: Answer and Explanation

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Question: 130

3. Water has the specific heat 4.186 kJ/kg•°C, a boiling point of 100°C, and a heat of vaporization of 2,260 kJ/kg. A sealed beaker contains 100 g of water that’s initially at 20°C. If the water absorbs 100 kJ of heat, what will its final temperature be?

A. 100°C
B. 119°C
C. 143°C
D. 183°C
E. 239°C

Correct Answer: A

Explanation:

First, let’s figure out how much heat is required to bring the water to its boiling point.

Q = mcT = (0.1 kg)(4.186 kJ/kg • °C)(100°C − 20°C) = 33 kJ

Once the water reaches 100°C, any additional heat will be absorbed and begin the transformation to steam. To completely vaporize the sample requires

Q = mL = (0.1 kg)(2260 kJ/kg) = 226 kJ

Since the 20°C water absorbed only 100 kJ of heat, enough heat was provided to bring the water to boiling, but not enough to completely vaporize it (at which point, the absorption of more heat would begin to increase the temperature of the steam). Thus, the water will reach and remain at 100°C.

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