Question

an electric heater supplies 37.4 joules of energy to 21.3 - g sample of h2o originally at 27.0 °c. compute the final temperature (in °c). current attempt in progress etextbook and media save for later attempts: 0 of 6 used submit answer

Explanation:

Step1: Identify the heat - capacity formula

The heat - energy formula is $Q = mc\Delta T$, where $Q$ is the heat energy, $m$ is the mass, $c$ is the specific heat capacity, and $\Delta T=T_f - T_i$ is the change in temperature. The specific heat capacity of water $c = 4.18\ J/(g\cdot^{\circ}C)$.

Step2: Rearrange the formula to solve for $\Delta T$

We can rewrite the formula as $\Delta T=\frac{Q}{mc}$. Given $Q = 37.4\ J$, $m = 21.3\ g$, and $c = 4.18\ J/(g\cdot^{\circ}C)$.
Substitute the values: $\Delta T=\frac{37.4\ J}{21.3\ g\times4.18\ J/(g\cdot^{\circ}C)}$.
First, calculate the denominator: $21.3\ g\times4.18\ J/(g\cdot^{\circ}C)=21.3\times4.18\ J/^{\circ}C = 88.934\ J/^{\circ}C$.
Then, $\Delta T=\frac{37.4\ J}{88.934\ J/^{\circ}C}\approx0.42^{\circ}C$.

Step3: Calculate the final temperature

We know that $\Delta T=T_f - T_i$, so $T_f=T_i+\Delta T$. Given $T_i = 27.0^{\circ}C$ and $\Delta T\approx0.42^{\circ}C$.
$T_f=27.0^{\circ}C + 0.42^{\circ}C=27.42^{\circ}C$.

Answer:

$27.42$