Question

1. to what temperature will a 50.0 g piece of glass raise if it absorbs 5275 joules of heat and its specific heat capacity is 0.5 j/g°c? the initial temperature of the glass is 20.0°c.

Explanation:

Step1: Recall the heat - transfer formula

The heat - transfer formula is $Q = mc\Delta T$, where $Q$ is the heat absorbed, $m$ is the mass, $c$ is the specific heat capacity, and $\Delta T$ is the change in temperature. We want to find $\Delta T$, so we can re - arrange the formula to $\Delta T=\frac{Q}{mc}$.

Step2: Substitute the given values

We are given that $Q = 5275\ J$, $m = 50.0\ g$, and $c=0.5\ J/g^{\circ}C$. Substituting these values into the formula $\Delta T=\frac{Q}{mc}$, we get $\Delta T=\frac{5275\ J}{50.0\ g\times0.5\ J/g^{\circ}C}$.

Step3: Calculate the change in temperature

First, calculate the denominator: $50.0\ g\times0.5\ J/g^{\circ}C = 25\ J/^{\circ}C$. Then, $\Delta T=\frac{5275\ J}{25\ J/^{\circ}C}=211^{\circ}C$.

Step4: Calculate the final temperature

The final temperature $T_f$ is related to the initial temperature $T_i$ and the change in temperature $\Delta T$ by the formula $T_f=T_i + \Delta T$. Given $T_i = 20.0^{\circ}C$ and $\Delta T = 211^{\circ}C$, then $T_f=20.0^{\circ}C+211^{\circ}C = 231^{\circ}C$.

Answer:

$231^{\circ}C$