Question

a chemical reaction takes place inside a flask submerged in a water bath. the water bath contains 5.50 kg of water at 24.3 °c. during the reaction 70.4 kj of heat flows out of the bath and into the flask.
calculate the new temperature of the water bath. you can assume the specific heat capacity of water under these conditions is 4.18 j·g⁻¹·k⁻¹. be sure your answer has the correct number of significant digits.

Explanation:

Step1: Convert mass to grams and heat to joules

Mass of water \( m = 5.50\space kg = 5500\space g \) (since \( 1\space kg = 1000\space g \)).
Heat lost by water bath \( q = - 70.4\space kJ = - 70400\space J \) (negative because heat is lost).

Step2: Use the heat formula \( q = mc\Delta T \) to find \( \Delta T \)

The formula for heat is \( q = mc\Delta T \), where \( c = 4.18\space J\cdot g^{-1}\cdot K^{-1} \) (specific heat capacity of water).
Rearranging for \( \Delta T \):
\( \Delta T = \frac{q}{mc} \)
Substitute values:
\( \Delta T = \frac{-70400\space J}{5500\space g \times 4.18\space J\cdot g^{-1}\cdot K^{-1}} \)
Calculate denominator: \( 5500 \times 4.18 = 22990 \)
\( \Delta T = \frac{-70400}{22990} \approx -3.06\space K \)

Step3: Relate temperature change in Kelvin to Celsius

A change in Kelvin is the same as a change in Celsius (\( \Delta T(K) = \Delta T(^\circ C) \)).
Initial temperature \( T_i = 24.3^\circ C \).
New temperature \( T_f = T_i + \Delta T \)
\( T_f = 24.3^\circ C - 3.06^\circ C = 21.24^\circ C \approx 21.2^\circ C \) (considering significant digits: 70.4 has 3, 5.50 has 3, 24.3 has 3; result should have 3 significant digits).

Answer:

\( 21.2^\circ C \)