© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Identify the Heat Gained and Lost
When the hot metal ball is dropped into the cooler water (and vessel), heat will flow from the metal ball to the water and vessel until thermal equilibrium is reached. We assume no heat loss to the surroundings.
Step 2: Define Symbols and Known Values
Mass of metal ball, $m_{\text{metal}} = 0.1 \,\text{kg}$
Specific heat of metal, $c_{\text{metal}} = 400 \,\text{J\,kg}^{-1}\,\text{K}^{-1}$
Initial temperature of metal ball, $T_{\text{metal, initial}} = 500\,^\circ\text{C}$
Mass of water, $m_{\text{water}} = 0.5\,\text{kg}$
Specific heat of water, $c_{\text{water}} = 4200\,\text{J\,kg}^{-1}\,\text{K}^{-1}$
Heat capacity of vessel, $C_{\text{vessel}} = 800\,\text{J\,K}^{-1}$
Initial temperature of water and vessel, $T_{\text{initial}} = 30\,^\circ\text{C}$
Final temperature (common equilibrium), $T_{\text{final}} = T$ (unknown)
Step 3: Write the Heat Balance Equation
According to the principle of calorimetry, total heat lost by the hot metal ball
= total heat gained by the water + vessel:
$m_{\text{metal}} \, c_{\text{metal}} \,\bigl[T_{\text{metal, initial}} - T\bigr] \;=\; m_{\text{water}} \, c_{\text{water}} \,\bigl[T - T_{\text{initial}}\bigr] \;+\; C_{\text{vessel}}\,(T - T_{\text{initial}})$.
Substituting the numbers:
$0.1 \times 400 \,\bigl(500 - T\bigr) \;=\; 0.5 \times 4200 \,\bigl(T - 30\bigr) \;+\; 800 \,\bigl(T - 30\bigr).$
Step 4: Simplify the Equation
Left side:
$0.1 \times 400 \,(500 - T) = 40\,(500 - T).$
Right side:
$0.5 \times 4200 \,(T - 30) + 800\,(T - 30) = 2100\,(T - 30) + 800\,(T - 30) = 2900\,(T - 30).$
So the equation becomes:
$40\,(500 - T) \;=\; 2900\,(T - 30).$
Step 5: Solve for the Final Temperature $T$
Expand both sides:
$20000 - 40T \;=\; 2900T - 87000.$
Rearrange:
$20000 + 87000 \;=\; 2900T + 40T = 2940T.$
Hence,
$107000 = 2940\,T \;\Longrightarrow\; T = \frac{107000}{2940} \approx 36.4\,^\circ\text{C}.$
Step 6: Calculate Percentage Increase in Water Temperature
The original water temperature was $30\,^\circ\text{C}$, and the new temperature is approximately $36.4\,^\circ\text{C}$.
Increase in temperature $= 36.4 - 30 = 6.4\,^\circ\text{C}.$
The percentage increment relative to the initial $30\,^\circ\text{C}$ is:
$$
\frac{6.4}{30} \times 100 \,\% \;\approx\; 21.3\% \,\approx\, 20\%.
$$
Final Answer
The approximate percentage increment in the temperature of the water is about 20%.
