Question

An insulated beaker with negligible mass contains liquid water with a mass of 0.270 kg and a temperature of 82.5 ∘C . How much ice at a temperature of -22.3 C must be dropped into the water so that the final temperature of the system will be 34.0 C?

Answer 1

Step-by-step explanation:

Given

mass of water
`m_1=0.27\ kg`

Temperature of water
`T_(wi)=82.5^(\circ)C`

Initial Temperature of ice
`=-22.3^(\circ)C`

Final temperature of system
`T=34^(\circ)C`

specific heat of water
`c=4.18\ kJ/kg-K`

specific heat of ice
`c_i=2.108\ kJ/kg-K`

Latent heat of ice
`L=336\ kJ/kg`

Heat loss by Water is equal to heat gained by ice

Heat loss by water
`Q_1=m_w* c* \Delta T`

`Q_1=0.27* 4.18* (82.5-34)=54.7371\ kJ`

Heat gained by ice
`Q_1=x* c_i(0-(-22.3))+x* L+x* c* (T-0)`

`Q_2=x* 2.108* (22.3)+x* 336+x* 4.18* 34`

`Q_2=525.1284x\ kJ`

`Q_1=Q_2`

`x=(54.73)/(525.1284)`

`x=0.104\ kg`