When 1460 mL of water at 260 Celicus is mixed with 510 mL of

When 146.0 mL of water at 26.0 Celicus is mixed with 51.0 mL of water at 85.0 celcius, what is the final temperature ( Assume that no heat is lost to the surrounding of water is 1.00 g/mL)

Solution

Given density of water = 1 gm/mL

146.0 mL water = 146 mL x 1 gm/mL = 146 gm at 26C & 51 gm at 85C

The energy amount going out of the warm water is equal to the energy amount going into the cool water. so

q_lost = q_gain

(mct) = - [(mct)] i.e. = (mc (Tf – T₁) = - [(mc(Tf – T₂)]

Specific heat of water (c) = 4.184 J/gC constant & T₁= 26 C, T₂= 85 C so we have

(146g)(4.184J/g•)(Tf - 26) = - [(51g)(4.184J/g•)(Tf - 85) ]

since 4.184 is common to both sides, it can be cancelled out. &

solve for Final temp ( Tf )

(146g)(Tf - 26) = - [(51g)( Tf - 85) ] i.e. (146 Tf – 3796 ) = -(51 Tf – 4335 )

146Tf – 3796 = 4335 – 51Tf

146 Tf + 51 Tf = 4335 + 3796

197 Tf = 8131 so

Tf = 41.27