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2 years ago

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A 15.0-gram lead ball at 25.0°C was heated with 40.5 joules of heat. Given the specific heat of lead is 0.128 J/g∙°C, what is th

e final temperature of the lead?

1 answer:

mr Goodwill [35]2 years ago

3 0

Answer:

$T=4985.5^{\circ}K$

Explanation:

The equation that relates heat Q with the temperature change $T-T_0$ of a substance of mass <em>m </em>and specific heat <em>c </em>is $Q=mc(T-T_0)$ .

We want to calculate the final temperature <em>T, </em>so we have:

$T=\frac{Q}{mc}+T_0$

Which for our values means (in this case we do not need to convert the mass to Kg since <em>c</em> is given in g also and they cancel out, but we add $273^{\circ}$ to our temperature in $^{\circ}C$ to have it in $^{\circ}K$ as it must be):

$T=\frac{Q}{mc}+T_0=\frac{40.5J}{(15g)(0.128J/g^{\circ}C)}+(298^{\circ}K)=4985.5^{\circ}K$

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