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A cube of gold weighing 192.4g is heated from 30 Celcius to some higher temperature, with the absorption of 226 joules of heat. The specific heat of gold is 0.030 J/g.C. WHat was the final temperature of the gold?

User Merkle Groot
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1 Answer

24 votes
24 votes

Answer:

69.2 °C

Step-by-step explanation:

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

Q = mcΔT

In this equation,

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

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

------> c = specific heat capacity (J/g°C)

------> ΔT = T₂ - T₁ = change in temperature (°C)

To clarify, "T₁" represents the initial temperature and "T₂" represents the final temperature. You can plug the given values into the equation and simplify to find "T₂".

Q = 226 J c = 0.030 J/g°C

m = 192.4 g ΔT = T₂ - T₁ = T₂ - 30°C

Q = mcΔT <----- Given equation

Q = mc(T₂ - T₁) <----- Rewrite ΔT

226 J = (192.4 g)(0.030 J/g°C)(T₂ - 30°C) <----- Insert variables

226 J = 5.772(T₂ - 30°C) <----- Multiply 192.4 and 0.030

39.154 = T₂ - 30°C <----- Divide both sides by 5.772

69.2 = T₂ <----- Add 30 to both sides

User Dave Sumter
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