A 5.0×101 kg sample of water absorbs 312 kJ of heat.If the water was initially at 20.1C, what is its final temperature?

Q=mc∆t

Q = heat energy (in joules) be sure to convert your kJ of heat to J by multiplying by 1000
m = mass (in grams)
be sure to convert your kg of water into grams by multiplying by 1000
c = specific heat (of water)
∆t= change in temperature (= final - initial temp)

c= 4.186 joule/gram °C

To find the final temperature of the water, we can use the formula:

q = mcΔT

where q is the heat absorbed by the water, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature.

In this case, we are given the mass of the water (5.0×10^1 kg), the heat absorbed by the water (312 kJ), and the initial temperature (20.1°C).

First, let's convert the given mass from kg to grams:
1 kg = 1000 g
Therefore, 5.0×10^1 kg = 5.0×10^4 g

Next, let's convert the heat absorbed from kJ to J:
1 kJ = 1000 J
Therefore, 312 kJ = 312000 J

The specific heat capacity of water is approximately 4.18 J/g°C.

Now, rearrange the formula to solve for ΔT:
ΔT = q / (mc)

Substitute the given values into the equation:
ΔT = 312000 J / (5.0×10^4 g × 4.18 J/g°C)

Simplify the equation:
ΔT = 1.490 J/°C

Finally, add ΔT to the initial temperature to find the final temperature:
Final temperature = Initial temperature + ΔT
Final temperature = 20.1°C + 1.490°C

Calculating the final temperature, we get:
Final temperature ≈ 21.59°C

Therefore, the final temperature of the water is approximately 21.59°C after absorbing 312 kJ of heat.