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Step-by-Step Solution
Step 1: Understand the Problem
A copper ball of mass 100 g, initially at an unknown temperature $T$, is dropped into a copper calorimeter (also made of copper) of mass 100 g. This calorimeter contains 170 g of water, all initially at 30°C (the room temperature). After mixing, the final equilibrium temperature of the system is 75°C. We need to find the initial temperature $T$ of the copper ball.
Step 2: Note the Given Data
Mass of copper ball, $m_{\text{ball}} = 100 \text{ g}$
Mass of calorimeter (copper), $m_{\text{cal}} = 100 \text{ g}$
Mass of water, $m_{\text{water}} = 170 \text{ g}$
Final equilibrium temperature of mixture, $T_{\text{final}} = 75^\circ \text{C}$
Initial temperature of calorimeter and water, $T_{\text{initial}} = 30^\circ \text{C}$
Specific heat capacity of copper, $c_{\text{Cu}} = 0.1 \text{ cal/(g·}^\circ \text{C)}$
Specific heat capacity of water, $c_{\text{water}} = 1 \text{ cal/(g·}^\circ \text{C)}$
Step 3: Apply the Principle of Calorimetry
According to the principle of calorimetry, the heat lost by the hotter object (copper ball) equals the heat gained by the cooler objects (copper calorimeter + water), assuming no heat loss to surroundings:
$ \text{Heat lost by copper ball} = \text{Heat gained by calorimeter} + \text{Heat gained by water} $
Step 4: Write the Heat Balance Equations
1. Heat lost by copper ball (mass $m_{\text{ball}}$, specific heat $c_{\text{Cu}}$, temperature change from $T$ to $75^\circ \text{C}$):
$ Q_{\text{lost}} = m_{\text{ball}} \, c_{\text{Cu}} \, (T - 75). $
2. Heat gained by copper calorimeter (mass $m_{\text{cal}}$, specific heat $c_{\text{Cu}}$, temperature change from $30^\circ \text{C}$ to $75^\circ \text{C}$):
$ Q_{\text{cal}} = m_{\text{cal}} \, c_{\text{Cu}} \, (75 - 30). $
3. Heat gained by water (mass $m_{\text{water}}$, specific heat $c_{\text{water}}$, temperature change from $30^\circ \text{C}$ to $75^\circ \text{C}$):
$ Q_{\text{water}} = m_{\text{water}} \, c_{\text{water}} \, (75 - 30). $
Step 5: Substitute Numerical Values
From the heat balance:
$ m_{\text{ball}} \, c_{\text{Cu}} \, (T - 75) = m_{\text{cal}} \, c_{\text{Cu}} \, (75 - 30) + m_{\text{water}} \, c_{\text{water}} \, (75 - 30). $
Substitute:
$ m_{\text{ball}} = 100 \text{ g}$
$ c_{\text{Cu}} = 0.1 \text{ cal/(g·}^\circ \text{C)}$
$ m_{\text{cal}} = 100 \text{ g}$
$ m_{\text{water}} = 170 \text{ g}$
$ 100 \times 0.1 \times (T - 75) = 100 \times 0.1 \times (75 - 30) + 170 \times 1 \times (75 - 30). $
Step 6: Simplify the Equation
Left side:
$ 100 \times 0.1 \times (T - 75) = 10 \,(T - 75). $
Right side:
$ 100 \times 0.1 \times 45 + 170 \times 45 = 10 \times 45 + 170 \times 45. $
$ 10 \times 45 = 450, \quad 170 \times 45 = 7650, $
so
$ 450 + 7650 = 8100. $
Step 7: Solve for $T$
Equating both sides:
$ 10 \,(T - 75) = 8100. $
$ T - 75 = \frac{8100}{10} = 810. $
$ T = 810 + 75 = 885. $
Step 8: Final Answer
The initial temperature $T$ of the copper ball is $885^\circ \text{C.}$
