Question

a 23.7 g piece of iron at 54.9°c is cooled and releases 338 j of heat. the specific heat of iron is 0.450 j/g°c. what is the final temperature of the iron? a 13.7°c b 27.9°c c 31.8°c d 23.2°c

Explanation:

Step1: Recall the heat - transfer formula

The formula for heat transfer is \(q = mc\Delta T\), where \(q\) is the heat, \(m\) is the mass, \(c\) is the specific heat capacity, and \(\Delta T=T_{final}-T_{initial}\). Since heat is released, \(q=- 338\space J\), \(m = 23.7\space g\), \(c=0.450\space J/g^{\circ}C\), and \(T_{initial}=54.9^{\circ}C\).

Step2: Substitute the values into the formula and solve for \(T_{final}\)

We have \(q = mc(T_{final}-T_{initial})\). Rearranging for \(T_{final}\) gives \(T_{final}=\frac{q}{mc}+T_{initial}\).
Substitute \(q=-338\space J\), \(m = 23.7\space g\), and \(c = 0.450\space J/g^{\circ}C\) into the formula:

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Answer:

d \(23.2^{\circ}C\)