Question

What temperature change in °C is produced when 4430 Joules are absorbed by 237 grams of water?

Answer 1

4.46 °C

Step-by-step explanation:

Step 1: Given and required data

• Energy absorbed (Q): 4430 J

• Mass of water (m): 237 g

• Specific heat of water (c): 4.186 J/g.°C

Step 2: Calculate the temperature change

We will use the following expression.

Q = c × m × ΔT

ΔT = Q / c × m

ΔT = 4430 J / (4.186 J/g.°C) × 237 g

ΔT = 4.46 °C

The temperature change is 4.46 °C.