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A pan containing 20.0 grams of water was allowed to cool from a temperature of 95.0 °C. If the amount of heat released is 1,200 joules, what is the approximate final temperature of the water?

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Answer:

80.7 °C

Explanation:

Step 1: Given and required data

  • Mass of water (m): 20.0 g
  • Initial temperature (T₁): 95.0 °C
  • Heat released (Q): -1,200 J (the negative sign is due to it being released)
  • Specific heat capacity of water (c): 4.184 J/g.°C

Step 2: Calculate the final temperature of the water

We will use the following expression.

Q = c × m × (T₂ - T₁)

T₂ = Q / c × m + T₁

T₂ = -1,200 J / (4.184 J/g.°C) × 20.0 g + 95.0 °C = 80.7 °C

Answer:

81 C

Explanation:

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