Question

a 90.0 g piece of metal, initially at 98.60 °c, is placed into 120.0 g of water initially at 24.30 °c. if the final temperature of the water is 34.00 °c, what is the specific heat of the metal? (the specific heat of water is 4.184 j/g·°c). answer: j/g·°c

Explanation:

Step1: Calculate heat absorbed by water

According to the formula \(Q = mc\Delta T\), where \(m = 120.0\ g\), \(c = 4.184\ J/g\cdot^{\circ}C\), \(\Delta T=34.00 - 24.30=9.70^{\circ}C\).
\(Q_{water}=120.0\times4.184\times9.70\)

$$
LATEXBLOCK0
$$

Step2: Calculate heat released by metal

Since \(Q_{metal}=-Q_{water}\) (heat lost by metal equals heat gained by water in an isolated system), \(Q_{metal}=- 4870.176\ J\). For the metal, \(m = 90.0\ g\), \(\Delta T=34.00 - 98.60=-64.60^{\circ}C\). Using the formula \(Q = mc\Delta T\) again, we can solve for \(c_{metal}\). Rearranging the formula gives \(c_{metal}=\frac{Q_{metal}}{m_{metal}\Delta T_{metal}}\)

$$
LATEXBLOCK1
$$

Answer:

\(0.838\ J/g\cdot^{\circ}C\)