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5.25g of unknown metal at 990C was placed into 15.0 g of water at 250C. The mixture came to a final temperature of 410C. What is the specific heat of the metal if the specific heat of water is 4.184 J/g0C?Group of answer choices191 J/gºC17.3 J/gºC3.30 J/gºC-3.30 J/gºC

Question

5.25g of unknown metal at 990C was placed into 15.0 g of water at 250C. The mixture came to a final temperature of 410C. What is the specific heat of the metal if the specific heat of water is 4.184 J/g0C?Group of answer choices191 J/gºC17.3 J/gºC3.30 J/gºC-3.30 J/gºC

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Solution

To solve this problem, we need to use the formula for heat transfer:

q = mcΔT

where: q = heat energy m = mass c = specific heat ΔT = change in temperature

We know that the heat lost by the metal is equal to the heat gained by the water. So, we can set up the following equation:

q(metal) = q(water)

Substituting the formula for heat transfer into both sides gives us:

m(metal)c(metal)ΔT(metal) = m(water)c(water)ΔT(water)

We can plug in the given values:

5.25g * c(metal) * (99ºC - 41ºC) = 15.0g * 4.184 J/gºC * (41ºC - 25ºC)

Solving for c(metal) gives us:

c(metal) = [15.0g * 4.184 J/gºC * (41ºC - 25ºC)] / [5.25g * (99ºC - 41ºC)]

After calculating the above expression, we find that the specific heat of the metal is approximately 3.30 J/gºC. So, the correct answer is 3.30 J/gºC.

This problem has been solved

Similar Questions

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