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:

 -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

8.7 degrees celsius and 15.6 degrees fahrenheit

Answer) by studying proton collisions hope this : )