Question

part a
how much heat is required to warm 1.70 l of water from 42.0 °c to 100.0 °c? (assume a density of 1.00 g/ml for the water. specific heat capacity of water is 4.184 j g⁻¹ °c⁻¹)
express your answer in joules to three significant figures.
q = j

Explanation:

Step1: Calculate the mass of water

First, use the density - volume formula $m=
ho V$. Given $
ho = 1.00\ g/mL$ and $V = 1.70\ L=1700\ mL$. So, $m=
ho V=1.00\ g/mL\times1700\ mL = 1700\ g$.

Step2: Calculate the heat required

Use the heat - transfer formula $q = mc\Delta T$, where $m = 1700\ g$, $c = 4.184\ J\ g^{-1}\ ^{\circ}C^{-1}$, and $\Delta T=T_f - T_i=100^{\circ}C - 42^{\circ}C = 58^{\circ}C$. Then $q=1700\ g\times4.184\ J\ g^{-1}\ ^{\circ}C^{-1}\times58^{\circ}C$.
$q = 1700\times4.184\times58\ J=1700\times242.672\ J = 412542.4\ J$.
Rounding to three significant figures, $q = 413000\ J$.

Answer:

$413000$