Question
Answer the above one
A calorimeter contains 542 mL of water at 40 degree C and 48 g of ice at 0 degree C. Find the final temperature of the system. The specific heat of water is 1 cal / g. degree C and the latent heat of fusion of water is 3. 33 x105 J / kg. Answer in units of degree C A calorimeter contains 510 mL of water at 56degree C and 24 g of ice at 0 degree C. Find the final temperature of the system. The specific heat of water is 1 cal / g. degree C and the latent heat of fusion of water is 3. 33 x 105 J / kg. Correct answer: 49. 9077degree C. Explanation: Let : V = 510 mL, mw = 510 g, mice = 0. 024 kg, Cw = 1 cal / g . degree C, Lf = 3. 33 x 105 J / kg, Tw = 56degree C, Ti = 0 degree C, and Delta T = 56degree C. The water cools to 0 degree C, giving off Qw = mw cw Delta T = (510 g)(1 cal / g . degree C)(56degree C) = 28560 cal heat. If all of the ice melts, it absorbs Qi = mice Lf = (0. 024 kg)(3. 33 x 105 J / kg) 1 cal / 4. 186 J = 1909. 29 cal heat. Since Qw > Qi1, all of the ice melts and also warms up to a temperature above 0 degree C. Thus Qlost water = Qmelt ice + Qgained mw Cw (Tw - Tf) = Qi + mice Cw (Tf - Tice) mw Cw Tw - mw Cw Tf = Qi + mice Cw (Tf - 0) mw Cw Tw - Qt = (mice Cw + mw CW) Tf Tf = mw Cw Tw- Qi / mice cw + mw cw. Tf 'mice Cw mw Since Mw Cw Tw - Qi = (510 g)(1 cal / g. degree C) x (56degree C) - 1909. 29 cal = 26650. 7 cal and mice Cw + mw cw = (0. 024 kg)(1 cal / g. degree C) + (510 g)(1 cal / g . degree C) = 534 cal / degree C, then Tf = 26650. 7 cal / 534 cal / degree = 49. 9077degree C.
Explanation / Answer
Let the final temperature by T.
Heat gained by ice=mL+mcT=0.048*333000+0.048*4186*T
Heat lost by water=mcT=0.542*4186*(40-T)
Conserving energy,
0.048*333000+0.048*4186*T=0.542*4186*(40-T)
Solving for T,
T=30.3C