Question

1. the heat capacity of solid iron is 0.447 j/g°c. if the same quantity of energy as in #3 were transferred to a 450 g chunk of iron at 20.°c, what would be the final temperature?

δ(65000 = 450g (0.447 j/g°c) δt
δ(65000 / 450g (0.447 j/g°c)) = δt
450g (0.447 j/g°c)

Explanation:

Step1: Recall the heat - transfer formula

The formula for heat transfer is \(Q = mc\Delta T\), where \(Q\) is the heat energy, \(m\) is the mass, \(c\) is the specific heat capacity, and \(\Delta T=T_{final}-T_{initial}\)

Step2: Rearrange the formula to solve for \(\Delta T\)

From \(Q = mc\Delta T\), we can get \(\Delta T=\frac{Q}{mc}\)

Step3: Substitute the given values

We know that \(Q = 65000J\), \(m = 450g\), and \(c=0.447J/g^{\circ}C\)

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Step4: Calculate the final temperature \(T_{final}\)

Since \(\Delta T=T_{final}-T_{initial}\) and \(T_{initial} = 20^{\circ}C\)

\(T_{final}=\Delta T+T_{initial}\)

\(T_{final}=323 + 20=343^{\circ}C\)

Answer:

\(343^{\circ}C\)