Question

part a
water (2310 g) is heated until it just begins to boil. if the water absorbs 5.83×10^5 j of heat in the process, what was the initial temperature of the water?
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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 of the substance, $c$ is the specific heat capacity, and $\Delta T$ is the change in temperature. For water, $c = 4.186\ J/(g\cdot^{\circ}C)$. The boiling point of water is $T_f=100^{\circ}C$. We want to find the initial temperature $T_i$. Rearranging the formula for $\Delta T$ gives $\Delta T=\frac{Q}{mc}$, and since $\Delta T=T_f - T_i$, then $T_i=T_f-\frac{Q}{mc}$.

Step2: Identify the given values

We have $m = 2310\ g$, $Q = 5.83\times10^{5}\ J$, $c = 4.186\ J/(g\cdot^{\circ}C)$ and $T_f = 100^{\circ}C$.

Step3: Calculate $\frac{Q}{mc}$

$\frac{Q}{mc}=\frac{5.83\times 10^{5}\ J}{2310\ g\times4.186\ J/(g\cdot^{\circ}C)}$
$=\frac{5.83\times 10^{5}}{2310\times4.186}\ ^{\circ}C$
$=\frac{5.83\times 10^{5}}{9679.66}\ ^{\circ}C\approx60.2^{\circ}C$

Step4: Calculate the initial temperature

$T_i=T_f-\frac{Q}{mc}=100^{\circ}C - 60.2^{\circ}C = 39.8^{\circ}C$

Answer:

$39.8^{\circ}C$