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Step-by-Step Solution
Step 1: Identify the Known Quantities
• Mass of water, $m = 200 \,\text{g} = 0.2 \,\text{kg}$
• Initial temperature, $T_{i} = 40^\circ C$
• Final temperature, $T_{f} = 60^\circ C$
• Change in temperature, $\Delta T = T_{f} - T_{i} = 60 - 40 = 20\,^\circ C$
• Specific heat capacity of water, $s = 4184\,\text{J\,kg}^{-1}\text{K}^{-1}$
Step 2: Note the Relevant Concept
For an isochoric (constant volume) process, the work done ($w$) is zero. Hence, from the first law of thermodynamics,
$ Q = \Delta U + w \quad \Rightarrow \quad Q = \Delta U \quad \text{(since } w = 0\text{)}.
$
Therefore, if the supplied heat $Q$ is used only to increase the internal energy, we have
$ \Delta U = m \, s \, \Delta T.
$
Step 3: Calculate the Change in Internal Energy
Substitute the known values into the equation:
$ \Delta U = 0.2\,\text{kg} \times 4184\,\text{J\,kg}^{-1}\text{K}^{-1} \times 20\,\text{K}.
$
Step 4: Perform the Numerical Computation
$ \Delta U = 0.2 \times 4184 \times 20 \,\text{J}
= 0.2 \times 4184 \times 20 \,\text{J}
= 16736 \,\text{J}
= 16.736 \,\text{kJ} \approx 16.7 \,\text{kJ}.
$
Step 5: State the Final Answer
Thus, the change in internal energy of the water is approximately $16.7 \,\text{kJ}$.
