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)`

Step-by-step 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* 10^5 J/kg`

Heat absorb by water when ice completely converts to water

`Q=mc_(ice)\delta T+mL`

`Q=0.045* 2.22* (0+10.2)+0.045* 3.4* 100`

Q=1.01898+15.3=16.3189 KJ

thus temperature of water will be

`3.95* 4.18* (20-T)=16.3189`

`T=19.011\approx 19^(\circ) C`