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What is the final temperature after 840 Joules is absorbed by 10.0g of water at 25.0oC?

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KIZUNAHIGHIN

Calculating Final Temperature.

Given

  • Heat Absorbed (q) = 840 J
  • Specfic Heat Capacity of water (c) = 4.184 J/g•°C
  • Initial Temperature (Δt) = 25°C
  • Mass (m) = 10.0 g
  • Final Temperature (F) = ?

Solve for the constant from the change of temperature.

Δt = q / mc

Δt = 840 J / (10.0g)(4.184 J / g•°C)

Δt = 840 J / 41.84 J / °C

Δt = 20.08°C

Solve for the final temperature

The final temperature will be the the sum of the two changes in temperature.

Δt = F - 25°

20.08 = F - 25

F = 20.08 + 25

F = 45.08

F ≈ 45.1 °C

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