A thermometer of mass 31.0g and specific heat 620 J/kg C degrees is initially at a temperature of 12.0 degrees C. If the thermometer is then immersed in a sample of water of mass 119g and the combination reaches a final temperature of 41.5 degrees C, what was the initial temperature of the water?
2007-07-24 03:52:38 · 4 answers · asked by night_gazer86 1 in Science & Mathematics ➔ Physics
The thermometer gained 567 J of heat from the water, so the water must have lost that much.
567 J = 620 * 0.031 kg * (41.5-12)
So the water lost 567 J...
-567 J = 119g * 4.18 J/(g C) * (41.5 C - ti)
Initial temperature = 42.64 degrees Celsius
2007-07-24 05:03:10 · answer #1 · answered by smilam 5 · 0⤊ 0⤋
Heat lost by thermometer...
= 0.031kg x 620J/kg/°C x 29.5°C ÎT = 567J
Heat gained by water ...
= (119g x 4.184J/g/°C) x (T - 41.5°C)
Heat lost = Heat gained.
567J = 498T - 20,662
20,662 + 567 = 498T
21,230 ÷ 498 = T
Water initial temp.(T) = 42.63°C
2007-07-24 12:18:00 · answer #2 · answered by Norrie 7 · 0⤊ 0⤋
It must be 46.26 degrees C, in my humble opinion.
2007-07-24 11:59:35 · answer #3 · answered by bmunavirov 1 · 0⤊ 0⤋
mc"delta" t = mc"delta" t
Solve for initail temp of water
2007-07-24 13:38:38 · answer #4 · answered by geezuskreyest 5 · 0⤊ 0⤋