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A 2.3 g sample of gold (specific heat capacity = 0.130 J/g °C) is heated using 92.3 J of energy. If the original temperature of the gold is 25.0°C, what is its final temperature?

User Darish
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1 Answer

4 votes

Answer:

325.38°C

Step-by-step explanation:

We can use the formula Q = mcΔT to solve this problem, where Q is the amount of heat transferred, m is the mass of the substance, c is its specific heat capacity, and ΔT is the change in temperature.

First, we need to calculate the amount of heat transferred:

Q = 92.3 J

Next, we need to calculate the change in temperature. We can rearrange the formula to solve for ΔT:

ΔT = Q / (mc)

where ΔT is the change in temperature, Q is the amount of heat transferred, m is the mass of the substance, and c is its specific heat capacity.

Plugging in the values we have:

ΔT = 92.3 J / (2.3 g x 0.130 J/g °C)

ΔT = 300.38 °C

This means that the temperature of the gold has increased by 300.38 °C. Since the initial temperature was 25.0°C, the final temperature will be:

Final temperature = Initial temperature + ΔT

Final temperature = 25.0°C + 300.38 °C

Final temperature = 325.38 °C

Therefore, the final temperature of the gold is 325.38°C.

User Jameshollisandrew
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