The specific heat of copper is 387 J/kg°C. The temperature of a 0.45-kg sample of copper decreases from 86.0 °C to 20.0 °C. How much heat flows out of the copper sample during this temperature drop? 8,700 J 9,600 J 11,000 J 14,000 J
Question
The specific heat of copper is 387 J/kg°C. The temperature of a 0.45-kg sample of copper decreases from 86.0 °C to 20.0 °C. How much heat flows out of the copper sample during this temperature drop? 8,700 J 9,600 J 11,000 J 14,000 J
Solution
To solve this problem, we need to use the formula for heat transfer which is:
q = mcΔT
where:
- q is the heat transferred,
- m is the mass of the substance,
- c is the specific heat capacity of the substance, and
- ΔT is the change in temperature.
Given:
- m = 0.45 kg,
- c = 387 J/kg°C,
- Initial temperature (Ti) = 86.0 °C,
- Final temperature (Tf) = 20.0 °C.
First, we need to find the change in temperature (ΔT), which is the final temperature (Tf) subtracted from the initial temperature (Ti). So, ΔT = Ti - Tf = 86.0 °C - 20.0 °C = 66.0 °C.
Then, we substitute the values into the formula:
q = mcΔT q = 0.45 kg * 387 J/kg°C * 66.0 °C
When you multiply these values together, you get q = 11,600 J.
So, the heat that flows out of the copper sample during this temperature drop is 11,600 J. The closest answer to this value is 11,000 J.
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