Question

mario and patience are making cookies. they are ready to drop balls of dough on baking sheets. they have baking sheets made of treated glass, aluminum, stainless steel, and copper. the specific heats of these substances are treated glass = 0.75 j/g·°c aluminum = 0.90 j/g·°c stainless steel = 0.50 j/g·°c copper = 0.39 j/g·°c based on the specific heat values, which sheet is going to warm up the quickest, and therefore, bake the cookies the fastest?

Explanation:

Step1: Recall the formula for heat

The formula for heat is \(Q = mc\Delta T\), where \(Q\) is the heat energy, \(m\) is the mass, \(c\) is the specific heat capacity, and \(\Delta T\) is the change in temperature.

Assuming the same mass \(m\) and the same amount of heat \(Q\) is absorbed (from the oven), we can rewrite the formula for \(\Delta T\) as \(\Delta T=\frac{Q}{mc}\).

If \(m\) and \(Q\) are constant, \(\Delta T\) is inversely proportional to \(c\) (specific - heat capacity). That is, \(\Delta T\propto\frac{1}{c}\)

Step2: Compare the specific - heat values

We are given the specific - heat values:

• Treated glass: \(c_{glass}=0.75\space J/g\cdot^{\circ}C\)
• Aluminum: \(c_{aluminum} = 0.90\space J/g\cdot^{\circ}C\)
• Stainless steel: \(c_{steel}=0.50\space J/g\cdot^{\circ}C\)
• Copper: \(c_{copper}=0.39\space J/g\cdot^{\circ}C\)

Since \(\Delta T\propto\frac{1}{c}\), the material with the smallest specific - heat capacity will have the largest \(\Delta T\) (will warm up the quickest for a given \(Q\) and \(m\))

Comparing the values \(0.39<0.50 < 0.75<0.90\)

Answer:

Copper