Question

6)

How many joules of heat energy must be added to 5 grams of liquid water to change its

temperature from 10°C to 30°C?

A)

83.6 joules

B)

100 joules

C)

312.6 joules

D)

418 joules

Answer 1

D) 418 J

Step-by-step 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