Question

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 the final temperature of the lead?

Answer 1

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 m and specific heat c is $Q=mc(T-T_0)$.

We want to calculate the final temperature T, 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 c 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):

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