A 45-kg sample of water absorbs 345 kJ of heat. If the water was initially at 22.1 degrees celcius, what was the final temperature?

To find the final temperature of the water, we can use the specific heat equation:

q = m * c * ΔT

Where:
- q is the heat absorbed or released by the substance
- m is the mass of the substance
- c is the specific heat capacity of the substance
- ΔT is the change in temperature

In this case, we are given the mass of the water (m = 45 kg), the heat absorbed (q = 345 kJ), and the initial temperature (22.1 degrees Celsius). We need to find the final temperature.

First, we need to convert the heat from kilojoules (kJ) to joules (J) because the specific heat capacity (c) is usually expressed in J/kg·°C.

345 kJ = 345,000 J

Now, let's rearrange the specific heat equation to solve for the change in temperature:

ΔT = q / (m * c)

The specific heat capacity of water is approximately 4.18 J/g·°C or 4,180 J/kg·°C.

ΔT = 345,000 J / (45 kg * 4,180 J/kg·°C)

ΔT = 1,533.07086505 °C

To find the final temperature, we need to add the change in temperature to the initial temperature:

Final temperature = Initial temperature + ΔT

Final temperature = 22.1 °C + 1,533.07086505 °C

Final temperature = 1555.17086505 °C

Therefore, the final temperature of the water is approximately 1555.17 °C.

q = mass x specific heat x (Tfinal-Tinitial)