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What is the final temperature in °C of 680.1 g of water (specific heat

= 4.184 J/g °C) at 24.20°C that absorbed 950.0 J of heat?

User Anjelika
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1 Answer

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Answer:

24.53 °C

Step-by-step explanation:

To find the final temperature, you need to use the following equation:

Q = mc(T₂ - T₁)

In this equation,

-----> Q = heat (J)

-----> m = mass (g)

-----> c = specific heat (J/g°C)

-----> T₂ = final temperature (°C)

-----> T₁ = initial temperature (°C)

Using the given values, you can rearrange the equation and simplify to find "T₂".

Q = 950.0 J T₂ = ? °C

m = 680.1 g T₁ = 24.20 °C

c = 4.184 J/g°C

Q = mc(T₂ - T₁) <----- Given equation

950.0 J = (680.1 g)(4.184 J/g°C)(T₂ - 24.20°C) <----- Insert values

950.0 J = 2845(T₂ - 24.20°C) <----- Multiply 680.1 and 4.184

0.3339 = T₂ - 24.20°C <----- Divide 950.0 by 2845

24.53 = T₂ <----- Add 24.20 to both sides

User Alexey Ferapontov
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