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: Recall the formula for heat energy

The formula to calculate the heat energy \( Q \) required to raise the temperature of a substance is \( Q = mc\Delta T \), where \( m \) is the mass of the substance, \( c \) is the specific heat capacity, and \( \Delta T \) is the change in temperature. For water, the specific heat capacity \( c = 4.18 \, \text{J/g}^\circ\text{C} \) (or \( 4.18 \, \text{kJ/kg}^\circ\text{C} \)).

Step2: Convert mass to kg

The mass of water \( m = 200 \, \text{g} = 0.2 \, \text{kg} \) (since \( 1 \, \text{kg} = 1000 \, \text{g} \)).

Step3: Calculate the change in temperature

The initial temperature \( T_1 = 27^\circ\text{C} \) and the final temperature \( T_2 = 100^\circ\text{C} \). So, \( \Delta T = T_2 - T_1 = 100 - 27 = 73^\circ\text{C} \).

Step4: Substitute values into the formula

Using \( Q = mc\Delta T \), with \( m = 0.2 \, \text{kg} \), \( c = 4.18 \, \text{kJ/kg}^\circ\text{C} \), and \( \Delta T = 73^\circ\text{C} \):

$$ Q = 0.2 \times 4.18 \times 73 $$

First, calculate \( 0.2 \times 4.18 = 0.836 \). Then, \( 0.836 \times 73 = 61.028 \approx 61.3 \, \text{kJ} \) (considering possible rounding in specific heat value or calculation steps, the closest option is 61.3 kJ).

Answer:

B. 61.3 kJ