Question

1. a 44.0 g sample of an unknown metal at 99.0°c was placed in a constant - pressure calorimeter containing 80.0 g of water at 24.0°c. the final temperature of the system was found to be 28.4°c. calculate the specific heat of the metal. (the specific heat capacity of water is 4.184 j/g°c.)

a. 2.03 j/g·°c
b. 1.01 j/g·°c
c. 0.474 j/g·°c
d. 0.246 j/g·°c
e. 3.22 j/g·°c

Explanation:

Step1: Calculate heat absorbed by water

Use formula \(q = mc\Delta T\).
\(m_{water}=80.0\ g\), \(c_{water} = 4.184\ J/g^{\circ}C\), \(\Delta T_{water}=28.4 - 24.0=4.4^{\circ}C\)
\(q_{water}=80.0\times4.184\times4.4\)
\(q_{water}=80.0\times18.4096 = 1472.768\ J\)

Step2: Calculate heat released by metal

Since \(q_{metal}=-q_{water}\) (heat lost by metal = heat gained by water)
\(q_{metal}=- 1472.768\ J\)
\(m_{metal}=44.0\ g\), \(\Delta T_{metal}=28.4 - 99.0=-70.6^{\circ}C\)
Use \(q = mc\Delta T\) to find \(c_{metal}\), \(c_{metal}=\frac{q_{metal}}{m_{metal}\Delta T_{metal}}\)
\(c_{metal}=\frac{-1472.768}{44.0\times(- 70.6)}\)
\(c_{metal}=\frac{-1472.768}{-3106.4}=0.474\ J/g^{\circ}C\)

Answer:

C. \(0.474\ J/g^{\circ}C\)