Question

a 50.3 g sample of polystyrene, which has a specific heat capacity of 1.880 j·g⁻¹·°c⁻¹, is put into a calorimeter (see sketch at right) that contains 200.0 g of water. the temperature of the water starts off at 19.0 °c. when the temperature of the water stops changing its 25.7 °c. the pressure remains constant at 1 atm. calculate the initial temperature of the polystyrene sample. be sure your answer is rounded to 2 significant digits.

Explanation:

Step1: Define heat transfer equation

Heat lost by polystyrene = Heat gained by water: $m_s c_s (T_{s,i} - T_f) = m_w c_w (T_f - T_{w,i})$

Step2: List known values

$m_s=50.3g, c_s=1.880J·g^{-1}·^\circ C^{-1}, m_w=200.0g, c_w=4.184J·g^{-1}·^\circ C^{-1}, T_{w,i}=19.0^\circ C, T_f=25.7^\circ C$

Step3: Calculate heat gained by water

$Q_w = 200.0×4.184×(25.7-19.0) = 200×4.184×6.7 = 5620.16J$

Step4: Solve for $T_{s,i}$

Rearrange equation: $T_{s,i} = T_f + \frac{Q_w}{m_s c_s} = 25.7 + \frac{5620.16}{50.3×1.880}$
Calculate denominator: $50.3×1.880≈94.564$
$\frac{5620.16}{94.564}≈59.43$
$T_{s,i}≈25.7+59.43≈85.13^\circ C$

Step5: Round to 2 sig figs

$85^\circ C$

Answer:

85