Question

A certain mass of water was heated with 41,840 Joules, raising its temperature from 22.0°C to 28.5 °C. Find the

mass of the water.

Answer 1

1.5 × 10³ g

Step-by-step explanation:

Step 1: Given and required data

• Transferred heat (Q): 41,840 J

• Initial temperature: 22.0 °C

• Final temperature: 28.5 °C

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

Step 2: Calculate the temperature change

ΔT = 28.5°C - 22.0 °C = 6.5 °C

Step 3: Calculate the mass (m) of water

We will use the following expression.

Q = c × m × ΔT

m = Q / c × ΔT

m = 41,840 J / (4.184 J/g.°C) × 6.5 °C = 1.5 × 10³ g