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Step-by-Step Solution
Step 1: Identify the Given Data
• Heat supplied, $q = 1.0\,\text{kJ} = 1000\,\text{J}$
• Molar heat capacity of water at constant pressure, $C = 75\,\text{J K}^{-1} \text{mol}^{-1}$
• Mass of water, $m = 100\,\text{g}$
• Molar mass of water, $M = 18\,\text{g mol}^{-1}$
Step 2: Calculate the Number of Moles of Water
Number of moles,
$$ n = \frac{m}{M} = \frac{100\,\text{g}}{18\,\text{g mol}^{-1}} = \frac{100}{18}\,\text{mol}. $$
Step 3: Use the Relation for Heat and Temperature Change
The relation between supplied heat ($q$), number of moles ($n$), molar heat capacity ($C$), and temperature change ($\Delta T$) is:
$$ q = n\,C\,\Delta T. $$
Step 4: Substitute the Values and Solve for $\Delta T$
Substituting the known values:
$$ 1000\,\text{J} = \left(\frac{100}{18}\,\text{mol}\right)\,(75\,\text{J K}^{-1} \text{mol}^{-1})\,\Delta T. $$
Therefore,
$$ \Delta T = \frac{1000\,\text{J}}{\left(\frac{100}{18} \times 75 \right)}. $$
First, compute the denominator:
$$ \frac{100}{18} \times 75 = \frac{100 \times 75}{18} = \frac{7500}{18}\,\text{J K}^{-1}. $$
So,
$$ \Delta T = \frac{1000}{\frac{7500}{18}} = \frac{1000 \times 18}{7500} = \frac{18000}{7500}. $$
Carrying out the division:
$$ \Delta T = 2.4\,\text{K}. $$
Step 5: Conclusion
The increase in temperature of the water is $2.4\,\text{K}.$
