Question

a 95.0 g sample of copper ($c_p = 0.20 j/°c·g$) is heated to 82.4°c and then placed in a container of water ($c_p = 4.18 j/°c·g$) at 22.0°c. the final temperature of the water and the copper is 25.1°c. what was the mass of the water in the original container? assume that all heat lost by the copper is gained by the water. (use the formulas below to help in your problem - solving).
$q_{metal} = q_{water}$
$-c_mm_mdelta t_m = c_wm_wdelta t_w$
\\(\circ\\) 0.240 g $h_2o$
\\(\circ\\) 4.73 g $h_2o$
\\(\circ\\) 84.0 g $h_2o$
\\(\circ\\) 36,700 g $h_2o$

Explanation:

Step1: Identify known values

For copper: \( m_{cu} = 95.0\ g \), \( c_{cu} = 0.20\ J/^{\circ}C\cdot g \), \( T_{i,cu} = 82.4^{\circ}C \), \( T_f = 25.1^{\circ}C \)
For water: \( c_{w} = 4.18\ J/^{\circ}C\cdot g \), \( T_{i,w} = 22.0^{\circ}C \), \( T_f = 25.1^{\circ}C \), \( m_w = ? \)

Step2: Calculate \( \Delta T \) for copper and water

\( \Delta T_{cu} = T_f - T_{i,cu} = 25.1 - 82.4 = -57.3^{\circ}C \)
\( \Delta T_{w} = T_f - T_{i,w} = 25.1 - 22.0 = 3.1^{\circ}C \)

Step3: Apply heat transfer formula \( -c_{cu}m_{cu}\Delta T_{cu} = c_{w}m_w\Delta T_{w} \)

Substitute values:
\( -(0.20)(95.0)(-57.3) = (4.18)(m_w)(3.1) \)
Calculate left side: \( (0.20)(95.0)(57.3) = 0.20 \times 95.0 \times 57.3 = 1088.7 \)
Right side: \( 4.18 \times 3.1 \times m_w = 12.958 \times m_w \)

Step4: Solve for \( m_w \)

\( m_w = \frac{1088.7}{12.958} \approx 84.0\ g \)

Answer:

84.0 g H₂O (corresponding to the option "84.0 g H₂O")