Question #44228

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Question #44228

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Answer 1 · 2015-12-02T00:29:09

$728 J$ $"728 J"$

Explanation:

All you have to do here is use the equation the establishes a relationship, in your case, between heat gained and increase in temperature

$q = m \cdot c \cdot Δ T$ $color(blue)(q = m * c * DeltaT)" "$ , where

$q$ $q$ - heat absorbed
$m$ $m$ - the mass of the sample
$c$ $c$ - the specific heat of the substance
$Δ T$ $DeltaT$ - the change in temperature, defined as final temperature minus initial temperature

So, you know that copper has a specific heat of

$c_{copper} = 0.385 \frac{J}{g^{\circ} C}$ $c_"copper" = 0.385"J"/("g" ""^@"C")$

So, what does a substance's specific heat tell you?

Well, it tells you how much heat is needed to increase the mass of a $1.00-g$ $"1.00-g"$ sample by $1^{\circ} C$ $1^@"C"$ . More specifically, you need to provide $0.385 J$ $"0.385 J"$ of heat to increase the mass of $1.00 g$ $"1.00 g"$ of copper by $1^{\circ} C$ $1^@"C"$ .

Now, if you have a bigger mass, you'd need more heat to increase its temperature by $1^{\circ} C$ $1^@"C"$ .

If you also want to increase its temperature by more than $1^{\circ} C$ $1^@"C"$ , you'd once again need more heat.

In your case, the change in temperature will be

$Δ T = {324.3}^{\circ} C - {20.5}^{\circ} C = {303.8}^{\circ} C$ $DeltaT = 324.3^@"C" - 20.5^@"C" = 303.8^@"C"$

This means that the amount of heat you'd need will be equal to

$q = 6.22 g \cdot 0.385 \frac{J}{g^{\circ} C} \cdot 303.8^{\circ} C = 727.51 J$ $q = 6.22 color(red)(cancel(color(black)("g"))) * 0.385"J"/(color(red)(cancel(color(black)("g"))) color(red)(cancel(color(black)(""^@"C")))) * 303.8color(red)(cancel(color(black)(""^@"C"))) = "727.51 J"$

Rounded to three sig figs, the answer will be

$q = 728 J$ $q = color(green)("728 J")$

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Question #44228

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