# find the final equilibrium temperature of water.

by user3538292   Last Updated October 20, 2019 05:20 AM - source

Would like to know the outcome if i place a 1 pound 1200°Fahrenheit rock into 1 gallon of water 50F. The formula i found was Q=m⋅c⋅ΔT, think this is what i need.

Specific heat of water is 4.186 joule/gram °C
Specific heat of limestone is 0.91 (kJ/(kg K)) as found below.
https://www.engineeringtoolbox.com/specific-heat-solids-d_154.html

1 pound of limestone = .460grams
1 gallon of water = 3.79kg

How i set up the equation.
https://socratic.org/questions/how-to-find-the-final-equilibrium-temperature-when-a-hot-iron-mass-is-placed-in-
Heat lost = Heat Gained
$${460\over 3790}$$ $$× .910 × (1200 - t)$$ = $$3.79 × 4.186 × (t - 50)$$
$$0.121 × .910 × (1200 - t)$$ = $$15.864 × (t - 50)$$
$$0.110 ×(1200 -t)$$ = $$15.864t - 793.2$$
$$132 - 0.110t$$ = $$15.864t - 793.2$$
$$925.2 = 15.974t$$
$$t = 57.91$$

I have no science background just curious because i watch a duel survivor episode in which they boiled water with a rock to disinfect. This answer seems really low and i think i might have messed up at the joule/gram °C and (kJ/(kg K) part since i am not sure if i need to convert. Sorry for not making the equations more math like, not computer smart.

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