Question

6)

Answer 1

Answer:

D) 418 J

Explanation:

Step 1: Given and required data

• Mass of water (m): 5 g

• Initial temperature: 10 °C

• Final temperature: 30 °C

• Specific heat capacity of water (c): 4.184 J/g.°C

Step 2: Calculate the temperature change (ΔT)

ΔT = 30 °C - 10 °C = 20 °C

Step 3: Calculate the heat required (Q)

We will use the following expression.

Q = c × m × ΔT

Q = 4.184 J/g.°C × 5 g × 20 °C = 418 J