Question

question 5 of 30 how much heat must be transferred to 3500 g of liquid water to change the water’s temperature from 27°c to 32°c? (the specific heat capacity of liquid water is 4.186 j/g·°c.)

a. 2900 j

b. 470,000 j

c. 170 j

d. 73,000 j

Explanation:

Step1: Recall the heat formula

The formula for heat transfer is \( Q = mc\Delta T \), where \( Q \) is heat, \( m \) is mass, \( c \) is specific heat capacity, and \( \Delta T \) is temperature change.

Step2: Calculate temperature change

\( \Delta T = T_2 - T_1 = 32^\circ\text{C} - 27^\circ\text{C} = 5^\circ\text{C} \)

Step3: Substitute values into formula

Given \( m = 3500\ \text{g} \), \( c = 4.186\ \text{J/g}^\circ\text{C} \), \( \Delta T = 5^\circ\text{C} \).
\( Q = 3500\ \text{g} \times 4.186\ \text{J/g}^\circ\text{C} \times 5^\circ\text{C} \)
First, calculate \( 3500 \times 5 = 17500 \), then \( 17500 \times 4.186 = 73255\ \text{J} \approx 73000\ \text{J} \)

Answer:

D. 73,000 J