Question

question 5 (1 point)
how much heat is gained when the temperature of 288 g of liquid water is raised from 100 °c to 125 °c?
30 618 j
30 168 j
20 168 j
20 618 j
120 384 j
you can use this periodic table and the enthalpies of formation and heat capacity reference tables during the quiz. however, as stated in the course outline, and in the orientation session, this quiz is not open book. you cannot access course material work as a group, use a

Explanation:

Step1: Recall the specific heat formula

The formula for heat \(Q = mc\Delta T\), where \(m\) is mass, \(c\) is specific heat capacity, and \(\Delta T\) is temperature change. For liquid water, \(c = 4.18\space J/g^{\circ}C\).

Step2: Calculate \(\Delta T\)

\(\Delta T=T_{final}-T_{initial}=125 - 100=25^{\circ}C\)

Step3: Substitute values into the formula

\(m = 288\space g\), \(c = 4.18\space J/g^{\circ}C\), \(\Delta T = 25^{\circ}C\)
\(Q=288\times4.18\times25\)
First, \(288\times25 = 7200\)
Then, \(7200\times4.18=30168\space J\)

Answer:

B. \(30168\space J\)