Question

question 14 of 17
the specific heat of a certain type of cooking oil is 0.418 cal/(g·°c). how much heat energy is needed to raise the temperature
of 0.548 kg of this oil from 23.0 °c to 60.0 °c?
q = cal

Explanation:

Step1: Convert mass from kg to g

Since \(1\ kg = 1000\ g\), for \(m = 0.548\ kg\), we have \(m=0.548\times1000 = 548\ g\).

Step2: Calculate the temperature change \(\Delta T\)

\(\Delta T=T_{final}-T_{initial}\). Given \(T_{initial}=23.0^{\circ}C\) and \(T_{final}=60.0^{\circ}C\), then \(\Delta T=60.0 - 23.0=37.0^{\circ}C\).

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

The specific heat \(c = 0.418\ cal/(g\cdot^{\circ}C)\), \(m = 548\ g\), \(\Delta T=37.0^{\circ}C\).
Substitute the values into the formula: \(q=(548\ g)\times(0.418\ cal/(g\cdot^{\circ}C))\times(37.0^{\circ}C)\).
First, calculate \(548\times0.418 = 229.064\), then \(229.064\times37.0=8475.368\).

Answer:

\(8475\) cal