exercise 3.86 - enhanced - with feedback <chapter 1 and 3 homework missed this? watch kcv 3.11, 3.12. read…
exercise 3.86 - enhanced - with feedback <chapter 1 and 3 homework missed this? watch kcv 3.11, 3.12. read sections 3.11, 3.12. you can click on the review link to access the section in your e text. a 5.0×10¹ kg sample of water absorbs 359 kj of heat. part a if the water was initially at 25.0 °c, what is its final temperature? tf =
Answer
Explanation:
Step1: Recall heat - energy formula
$Q = mc\Delta T$, where $Q$ is heat energy, $m$ is mass, $c$ is specific - heat capacity, and $\Delta T=T_f - T_i$. The specific - heat capacity of water $c = 4.184\ J/(g\cdot^{\circ}C)$. First, convert the mass of water to grams: $m = 5.0\times10^{3}\ g$ (since $1\ kg=1000\ g$) and $Q = 359\times10^{3}\ J$ (since $1\ kJ = 1000\ J$), $T_i=25.0^{\circ}C$.
Step2: Rearrange the formula for $\Delta T$
$\Delta T=\frac{Q}{mc}$. Substitute the values: $m = 5.0\times10^{3}\ g$, $c = 4.184\ J/(g\cdot^{\circ}C)$, and $Q = 359\times10^{3}\ J$ into the formula. $\Delta T=\frac{359\times 10^{3}\ J}{5.0\times10^{3}\ g\times4.184\ J/(g\cdot^{\circ}C)}$.
Step3: Calculate $\Delta T$
$\Delta T=\frac{359\times 10^{3}}{5.0\times10^{3}\times4.184}=\frac{359}{5.0\times4.184}=\frac{359}{20.92}\approx17.2^{\circ}C$.
Step4: Calculate the final temperature $T_f$
Since $\Delta T=T_f - T_i$, then $T_f=\Delta T + T_i$. Substitute $\Delta T = 17.2^{\circ}C$ and $T_i = 25.0^{\circ}C$ into the formula. $T_f=17.2^{\circ}C+25.0^{\circ}C = 42.2^{\circ}C$.
Answer:
$42.2$