Question

a 58.4 g sample of glass, which has a specific heat capacity of 0.670 j·g⁻¹·°c⁻¹, is put into a calorimeter (see sketch at right) that contains 250.0 g of water. the temperature of the water starts off at 19.0 °c. when the temperature of the water stops changing its 21.9 °c. the pressure remains constant at 1 atm. calculate the initial temperature of the glass sample. be sure your answer is rounded to the correct number of significant digits.

Explanation:

Step1: Calculate heat gained by water

The formula for heat \(q = mc\Delta T\). For water, \(m = 250.0\ g\), \(c = 4.184\ J\cdot g^{-1}\cdot^{\circ}C^{-1}\), \(\Delta T=T_{final}-T_{initial}=21.9 - 19.0=2.9^{\circ}C\).

$$q_{water}=250.0\times4.184\times2.9$$

$$q_{water}=250.0\times12.1336$$

$$q_{water}=3033.4\ J$$

Step2: Calculate initial temperature of glass

Since \(q_{glass}=-q_{water}\) (heat lost by glass = heat gained by water). For glass, \(m = 58.4\ g\), \(c = 0.670\ J\cdot g^{-1}\cdot^{\circ}C^{-1}\), \(T_{final} = 21.9^{\circ}C\), and \(q_{glass}=mc(T_{final}-T_{initial})\).

$$ - 3033.4=58.4\times0.670\times(21.9 - T_{initial})$$

First, calculate \(58.4\times0.670 = 39.128\)

$$ - 3033.4=39.128\times(21.9 - T_{initial})$$

Then, \(\frac{-3033.4}{39.128}=21.9 - T_{initial}\)

$$ - 77.53=21.9 - T_{initial}$$

$$T_{initial}=21.9 + 77.53$$

$$T_{initial}=99.43^{\circ}C\approx99.4^{\circ}C$$

Answer:

\(99.4^{\circ}C\)