Question

determine the final temperature of a 37.53 ml salt solution initially at 22.54°c when 4.76 kj of heat is added to the solution. the specific heat of the solution is 3.108 j/(°c·g) and the density of the solution is 1.109 g/ml. report your answer in units of degrees celsius to 4 sig figs, but do not include units in your answer.

Explanation:

Step1: Calculate the mass of the solution

Use the formula \(m=
ho V\). Given \(V = 37.53\space mL\) and \(
ho=1.109\space g/mL\).
\(m=(1.109\space g/mL)\times(37.53\space mL)=41.62077\space g\)

Step2: Convert the heat added to joules

Given \(q = 4.76\space kJ\). Since \(1\space kJ=1000\space J\), then \(q = 4.76\times1000\space J = 4760\space J\)

Step3: Use the heat - transfer formula \(q = mc\Delta T\) to find \(\Delta T\)

The formula \(\Delta T=\frac{q}{mc}\). Substitute \(q = 4760\space J\), \(m = 41.62077\space g\), and \(c=3.108\space J/(^{\circ}C\cdot g)\)
\(\Delta T=\frac{4760\space J}{41.62077\space g\times3.108\space J/(^{\circ}C\cdot g)}\)
\(\Delta T=\frac{4760}{129.45735}\space^{\circ}C\approx36.77\space^{\circ}C\)

Step4: Calculate the final temperature \(T_f\)

Use the formula \(T_f=T_i+\Delta T\). Given \(T_i = 22.54\space^{\circ}C\)
\(T_f=22.54\space^{\circ}C + 36.77\space^{\circ}C=59.31\space^{\circ}C\)

Answer:

59.31