Question

a large punch bowl holds 3.95 kg of lemonade (which essentially is water) at 20.0 degrees celsius. A 45 gram ice cube at -10.2 degrees celsius is placed in the lemonade. What is the final temperature of the system? (ignore any heat exchange with the bowl)

Answer 1

Answer:$19 ^{\circ}$

Explanation:

Given

mass of water $(m_w) 3.95 kg$

Temperature of water is $20^{\circ} C$

mass of ice$(m_i) =45 gm$

Temperature of ice $=-10.2 ^{\circ} C$

specific heat of water $4.18 kJ/kg-K$

specific heat of ice $2.22 kJ/kg-K$

Latent heat of fusion(L) $3.4\times 10^5 J/kg$

Heat absorb by water when ice completely converts to water

$Q=mc_{ice}\delta T+mL$

$Q=0.045\times 2.22\times (0+10.2)+0.045\times 3.4\times 100$

Q=1.01898+15.3=16.3189 KJ

thus temperature of water will be

$3.95\times 4.18\times (20-T)=16.3189$

$T=19.011\approx 19^{\circ} C$