Question

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the heat capacity of silver is 25.351 j mol⁻¹ °c⁻¹. if 5.00 mol of silver are heated by 2.00 °c, how much heat was transferred from the surroundings to the silver?
○ 254 j
○ 50.7 j
○ -127 j
○ -50.7 j
○ -254 j

Explanation:

Step1: Recall the formula for heat transfer

The formula for heat transfer (\(q\)) using molar heat capacity (\(C\)) is \(q = n \times C \times \Delta T\), where \(n\) is the number of moles, \(C\) is the molar heat capacity, and \(\Delta T\) is the change in temperature.

Step2: Identify the given values

We have \(n = 5.00\space mol\), \(C = 25.351\space J\space mol^{-1}\space ^\circ C^{-1}\), and \(\Delta T = 2.00\space ^\circ C\).

Step3: Substitute the values into the formula

\(q = 5.00\space mol\times25.351\space J\space mol^{-1}\space ^\circ C^{-1}\times2.00\space ^\circ C\)
First, multiply \(5.00\) and \(2.00\): \(5.00\times2.00 = 10.0\)
Then, multiply by \(25.351\): \(10.0\times25.351 = 253.51\space J\approx254\space J\)
Since the silver is being heated, heat is transferred from the surroundings to the silver, so \(q\) is positive.

Answer:

254 J