Question

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 in an insulated container \(Q_{metal}=-Q_{water}\) (heat lost by metal = heat gained by water). For metal, \(m_{metal} = 50\space g\), \(\Delta T_{metal}=25 - 90=- 65^{\circ}C\), and \(Q_{metal}=m_{metal}c_{metal}\Delta T_{metal}\)
\(2090=50\times c_{metal}\times(- 65)\)
\(c_{metal}=\frac{2090}{50\times(- 65)}\) (but we take magnitude as we are interested in specific heat capacity value)
\(c_{metal}=\frac{2090}{50\times65}=\frac{2090}{3250}\approx0.64\space J/g^{\circ}C\)

Answer:

\(0.64\)