Question

How many joules are absorbed when 83.2 grams of water is heated from 30.0 °C to 74.0 °C?

a
10,500 joules

b
12,200 joules

c
15,300 joules

d
18,200 joules

Answer 1

Step-by-step explanation:

To calculate the amount of heat absorbed when heating water, you can use the formula:

Q = mcΔT

Where:

- Q is the heat absorbed (in joules)

- m is the mass of the substance (in grams)

- c is the specific heat capacity of the substance (in J/g°C)

- ΔT is the change in temperature (in °C)

Given:

- Mass of water, m = 83.2 grams

- Initial temperature, T_initial = 30.0°C

- Final temperature, T_final = 74.0°C

- Specific heat capacity of water, c = 4.18 J/g°C (approximately)

Now, calculate the change in temperature:

ΔT = T_final - T_initial = 74.0°C - 30.0°C = 44.0°C

Now, plug the values into the formula:

Q = (83.2 g) * (4.18 J/g°C) * (44.0°C)

Q ≈ 15,311.84 joules

So, the closest option is:

c) 15,300 joules