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A 50.0-g sample of aluminum pellets (specific heat capacity = 0.89 J/g°C) and a 10.0-g sample of iron pellets (specific heat capacity = 0.45 J/g°C) are heated to 100.0°C. What is the final temperature of the metal and water mixture when dropped into 97.3 g of water at 22.0°C?

A. 38.4°C
B. 41.7°C
C. 47.2°C
D. 52.9°C

Answer :

Final answer:

To find the final temperature of the metal and water mixture, use the principle of heat transfer. Set up an equation where the heat lost by the metal is equal to the heat gained by the water. Solve the equation to find the final temperature. The correct answer is 52.9°C.

Explanation:

To find the final temperature of the metal and water mixture, we can use the principle of heat transfer. The heat lost by the metal will be gained by the water. We can use the equation:

qmetal + qwater = 0,

where qmetal and qwater are the amounts of heat lost by the metal and gained by the water, respectively.

Since there is no heat transfer to the surroundings, we can assume that:

qmetal = -qwater.

Using the formula for heat transfer:

q = mcΔT,

where q is the amount of heat transferred, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.

We plug in the given values:

  • For aluminum: qmetal = -50.0g * 0.89 J/g°C * (100.0°C - final temperature)
  • For iron: qmetal = -10.0g * 0.45 J/g°C * (100.0°C - final temperature)
  • For water: qwater = 97.3g * 4.18 J/g°C * (final temperature - 22.0°C)

Simplifying the equation gives:

-50.0g * 0.89 J/g°C * (100.0°C - final temperature) - 10.0g * 0.45 J/g°C * (100.0°C - final temperature) = 97.3g * 4.18 J/g°C * (final temperature - 22.0°C)

Solving this equation will give us the final temperature of the metal and water mixture. The correct answer is option d) 52.9°C.

Learn more about heat transfer here:

https://brainly.com/question/13433948

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