Use Σq = 0 = (m∙c∙ΔT)water + (m∙c∙ΔT)metal
Water
m = 125ml = 125g <=> (1ml H₂O = 1gm H₂O)
c = 1cal/g∙⁰C or 4.184j/g∙⁰C
ΔT = 25.4⁰C – 22.0⁰C
Metal
m = 2.36102g
c = ?
ΔT = 99.5⁰C – 22.0⁰C
(m∙c∙ΔT)water + (m∙c∙ΔT)metal = 0
[(125)(1)(25.4 – 22.0)] + [(2.36102)(c)(25.4 – 99.5)] = 0
Solve for ‘c’.
Can someone help me with this problem.... i have an exam coming tomorrow and im doing practice problems but i dont know how to do this
Calculate the specific heat of a metal if 2.36102 grams of it at 99.5 degrees celsius is added to 125.0 mL of water at 22.0 degrees celsius. The final temperature of the system is 25.4 degrees celsius.
2 answers
Oops! ΔT(metal) = 25.4⁰C – 99.5⁰C
1 answer