Question

calculating specific heat
a 50 g sample of an unknown metal is heated to 90.0°c. it is placed in a perfectly insulated container along with 100 g of water at an initial temperature of 20°c. after a short time, the temperature of both the metal and water become equal at 25°c. the specific heat of water is 4.18 j/g°c in this temperature range. what is the specific heat capacity of the metal?
record your answer with two significant figures. j/g°c

Explanation:

Step1: Calculate heat gained by water

The formula for heat \(Q = mc\Delta T\). For water, \(m = 100\space g\), \(c = 4.18\space J/g^{\circ}C\), \(\Delta T=25 - 20=5^{\circ}C\).
\(Q_{water}=m_{water}c_{water}\Delta T_{water}=100\times4.18\times5 = 2090\space J\)

Step2: Calculate heat lost by metal

Since the container is insulated, \(Q_{metal}=-Q_{water}\). For the metal, \(m_{metal} = 50\space g\), \(\Delta T_{metal}=25 - 90=- 65^{\circ}C\), and \(Q_{metal}=m_{metal}c_{metal}\Delta T_{metal}\)

Step3: Solve for \(c_{metal}\)

We know \(Q_{metal}=-Q_{water}\), so \(m_{metal}c_{metal}\Delta T_{metal}=- 2090\). Substitute \(m_{metal} = 50\space g\) and \(\Delta T_{metal}=-65^{\circ}C\)
\(c_{metal}=\frac{-Q_{water}}{m_{metal}\Delta T_{metal}}=\frac{-(- 2090)}{50\times65}=\frac{2090}{3250}\approx0.64\space J/g^{\circ}C\)

Answer:

\(0.64\space J/g^{\circ}C\)