Question

question 7 (1 point)
use the following information to answer the next question.
a tank contains 200 g of water maintained at temperature of 27 ° c. the water is heated to raise its temperature to 100 ° c. the heat energy required is x kj.
the value of x is
83.6 kj
61.3 kj
62.3 kj
60.2 kj
63.2 kj

Explanation:

Step1: Convert mass to kg

Given mass of water \(m = 200\space g=0.2\space kg\)

Step2: Determine temperature change

Initial temperature \(T_1 = 27^{\circ}C\), final temperature \(T_2 = 100^{\circ}C\). Temperature change \(\Delta T=T_2 - T_1=100 - 27 = 73^{\circ}C\)

Step3: Use heat - energy formula

The specific heat capacity of water \(c = 4.18\space kJ/(kg\cdot^{\circ}C)\). The formula for heat energy \(Q=mc\Delta T\). Substitute \(m = 0.2\space kg\), \(c = 4.18\space kJ/(kg\cdot^{\circ}C)\) and \(\Delta T = 73^{\circ}C\) into the formula:

$$
LATEXBLOCK0
$$

Answer:

\(61.3\space kJ\)