What is the amount of heat energy released when 50.0 grams of water is cooled from 20.0°C to 10.0°C?

Q = mc Δt

q = heat, m = mass (in grams), c=specific heat (4.18 J/°C * g or J/K * g for water), Δt = change in temperature (in kelvin)

For this equation:

q = ?, m = 50.0 g, c = 4.18 J/°C * g = 4.18 J/K * g, initial t = 20.0°C, final t = 10°C

To convert from °C to K, add 273 to °C

0°C = 273 K

Initial t = 20.0°C + 273 = 293 K

Final t = 10.0°C + 273 = 283 K

ΔT = final temperature - initial temperature = 283-293 = - 10 K

Plug in what you know:

q = (50.0 g) x (4.18 J/K * g) x (-10 K) = - 2090

Check sig figs and cancel out units:

Answer: - 2090 J (you need 3 sig figs from the original numbers given and the units are Joules since grams and Kelvin cancel out)

Check to make sure answer makes sense:

Since heat energy is released, the answer should be negative. The answer is negative and therefore, makes sense