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Answer :
209
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Rewritten by : Barada
The initial temperature of the limestone was [tex]\( 209 \°C} \)[/tex](expressed to three significant figures).
To find the initial temperature of the limestone, we can use the principle of conservation of energy. The heat lost by the limestone will be equal to the heat gained by the water.
First, let's denote the specific heat capacities, masses, and temperatures:
Specific heat capacity of water = 4.186 J/g°C
Mass of water = 75.0 g
Initial temperature of water = 23.1°C
The final temperature of both water and limestone = 51.9°C
The heat gained by the water [tex](\( q_w )[/tex]) is given by:
[tex]\[ q_w = m_w \cdot c_w \cdot (T_f - T_{i,w}) \][/tex]
The heat lost by the limestone [tex](\( q_l \))[/tex] is given by:
[tex]\[ q_l = m_l \cdot c_l \cdot (T_{i,l} - T_f) \][/tex]
Since the heat lost by the limestone is equal to the heat gained by the water, we have:
[tex]\[ q_l = q_w \][/tex]
Substituting the expressions for [tex]\( q_w \)[/tex] and [tex]\( q_l \)[/tex]:
[tex]\[ m_l \cdot c_l \cdot (T_{i,l} - T_f) = m_w \cdot c_w \cdot (T_f - T_{i,w}) \][/tex]
Plugging in the given values:
[tex]\[ 62.6 \cdot 0.921 \cdot (T_{i,l} - 51.9) = 75.0 \cdot 4.186 \cdot (51.9 - 23.1) \][/tex]
Calculate the right-hand side:
[tex]\[ 75.0 \cdot 4.186 \cdot 28.8 = 9045.84 \, \text{J} \][/tex]
Now solve for [tex]\( T_{i,l} \)[/tex]:
[tex]\[ 62.6 \cdot 0.921 \cdot (T_{i,l} - 51.9) = 9045.84 \][/tex]
[tex]\[ 57.6846 \cdot (T_{i,l} - 51.9) = 9045.84 \][/tex]
[tex]\[ T_{i,l} = 156.8 + 51.9 \][/tex]
[tex]\[ T_{i,l} = 208.7 \][/tex]
