Could someone SHOW me the steps to working this one out.
Assume no heat is lost to the surroundings, d of water is 1 g/mol
3 Answers
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155ml(1g/ml) = 155g H2O
75ml(1g/ml) = 75g H2O
(mc∆t) = – [(mc∆t)]
(155g)(4.184J/g•℃)(tf – 26℃) = – [(75g)(4.184J/g•℃)(tf – 85℃)
648.52tf – 16861.52 = -[313.8tf – 26673]
962.32tf = 43534.52
tf = 45℃
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I forget, so I might be wrong about this. Use this equation:
Q1 = Q2
m1Cp(Tf-T1) = m2Cp(T2-Tf)
m1(Tf-T1) = m2(T2-Tf) Cp cancels on both sides
where
m1 and m2 are the masses of the different waters, so you will need the density of water to determine this: d = m/V. You have both volumes:
V1 = 155 mL
V2 = 75 mL
T1 = 26 C
T2 = 85 C
Solve for Tf = final temperature.
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