Question

c. $e_{a(\text{reverse})} = e_{a(\text{forward})}$
d. $e_{\text{reverse}} = -delta h$
e. $e_{a(\text{reverse})} = e_{a(\text{forward})} + delta h$

1. a 50.0g sample of an unknown metal at 100.00°c is placed in 100.0ml of water at 25°c. the final temperature is 28.5°c. what is the specific heat capacity of the metal?

a. $2.44 \text{j/g·°c}$
b. $0.12 \text{j/g·°c}$
c. $0.41 \text{j/g·°c}$
d. $-0.41 \text{j/g·°c}$
e. $0.82 \text{j/g·°c}$

Explanation:

Step1: Recall heat transfer formula

The heat lost by the metal (\(q_{metal}\)) is equal to the heat gained by the water (\(q_{water}\)) (assuming no heat loss to the surroundings), so \(q_{metal} = -q_{water}\). The formula for heat is \(q = mc\Delta T\), where \(m\) is mass, \(c\) is specific heat capacity, and \(\Delta T\) is the change in temperature.

Step2: Determine mass of water

The density of water is \(1.00\ g/mL\), so for \(100.0\ mL\) of water, \(m_{water} = 100.0\ mL\times1.00\ g/mL = 100.0\ g\). The specific heat capacity of water (\(c_{water}\)) is \(4.184\ J/g\cdot^{\circ}C\).

Step3: Calculate \(\Delta T\) for water and metal

For water: \(\Delta T_{water} = 28.5^{\circ}C - 25.0^{\circ}C = 3.5^{\circ}C\)
For metal: \(\Delta T_{metal} = 28.5^{\circ}C - 100.0^{\circ}C = -71.5^{\circ}C\)

Step4: Set up heat equations

\(q_{metal} = m_{metal}c_{metal}\Delta T_{metal}\)
\(q_{water} = m_{water}c_{water}\Delta T_{water}\)
Since \(q_{metal} = -q_{water}\), we have:
\(m_{metal}c_{metal}\Delta T_{metal} = -m_{water}c_{water}\Delta T_{water}\)

Step5: Solve for \(c_{metal}\)

Plug in the values:
\(50.0\ g\times c_{metal}\times(-71.5^{\circ}C) = -100.0\ g\times4.184\ J/g\cdot^{\circ}C\times3.5^{\circ}C\)
Simplify the right side: \(-100.0\times4.184\times3.5 = -1464.4\ J\)
Left side: \(50.0\times(-71.5)\times c_{metal} = -3575\ g\cdot^{\circ}C\times c_{metal}\)
So, \(-3575\ c_{metal} = -1464.4\)
Divide both sides by \(-3575\):
\(c_{metal} = \frac{1464.4}{3575} \approx 0.41\ J/g\cdot^{\circ}C\)

Answer:

c. \(0.41\ J/g\cdot^{\circ}C\)