How many kg of ice must be dropped to make Tf= 22.7C?

[madison222s]

Homework Statement

An insulated beaker with neglible mass contains .3kg of water at a temp of 78.6 C. How many kg of ice at -19.6 C must be dropped into the water to make Tf= 22.7C?
Specific heat for water= 4190 J/kgK
specific heat for ice= 2100 J/kgK
heat of fusion for water= 334kJ/kg

Homework Equations

Q=mc(Tf-Ti)
Q=mL

The Attempt at a Solution

Q₁ = bring ice to 0 C

• Q=m_icec(Tf-Ti)
• Q=m_ice(2100 J/kgK)(0-19.6)
• Q= 41160m_ice

Q₂ = melt ice (phase change)

• Q=mL
• Q=m_ice(334000 J/kg)

Q₃= cool down water

• Q=m_waterc(Tf-Ti)
• Q=(.3kg)(4190 J/kgK)(22.7 - 78.6)
• Q=-70266.9 J

Q₁+Q₂+Q₃=0
41160m_ice + 334000m_ice + (-70266.9)=0
375160m_ice= 70266.9
m_ice = .187kg

This answer isn't correct. What did I do wrong or what did I miss? I'm thinking maybe I missed another Q but I don't know what it could be.

 

[SteamKing]

Homework Statement

An insulated beaker with neglible mass contains .3kg of water at a temp of 78.6 C. How many kg of ice at -19.6 C must be dropped into the water to make Tf= 22.7C?
Specific heat for water= 4190 J/kgK
specific heat for ice= 2100 J/kgK
heat of fusion for water= 334kJ/kg

Homework Equations

Q=mc(Tf-Ti)
Q=mL

The Attempt at a Solution

Q₁ = bring ice to 0 C

• Q=m_icec(Tf-Ti)
• Q=m_ice(2100 J/kgK)(0-19.6)
• Q= 41160m_ice

Q₂ = melt ice (phase change)

• Q=mL
• Q=m_ice(334000 J/kg)

Q₃= cool down water

• Q=m_waterc(Tf-Ti)
• Q=(.3kg)(4190 J/kgK)(22.7 - 78.6)
• Q=-70266.9 J

Q₁+Q₂+Q₃=0
41160m_ice + 334000m_ice + (-70266.9)=0
375160m_ice= 70266.9
m_ice = .187kg

This answer isn't correct. What did I do wrong or what did I miss? I'm thinking maybe I missed another Q but I don't know what it could be.

Remember, once the ice melts, the liquid from this mixes with the liquid already in the beaker.

The final temperature of 22.7 °C must be reached by all of the liquid in the beaker.

 

[madison222s]

Remember, once the ice melts, the liquid from this mixes with the liquid already in the beaker.

The final temperature of 22.7 °C must be reached by all of the liquid in the beaker.

Then would Q₃ have to include both the masses?

 

[SteamKing]

Then would Q₃ have to include both the masses?

Yep. You can't separate the melt water from the warm water already in the beaker when the ice was dropped in.

 

[madison222s]

Yep. You can't separate the melt water from the warm water already in the beaker when the ice was dropped in.

So then its
Q₃=(m_ice+m_water)c(Tf-Ti)
=(.3kg + m_ice)(4190)(22.7-78.6)
=(.3kg + m_ice)(-234221)
= -70266.3 - 234221m_ice

And then add this Q₃ to the others and set =0?

 

[SteamKing]

So then its
Q₃=(m_ice+m_water)c(Tf-Ti)
=(.3kg + m_ice)(4190)(22.7-78.6)
=(.3kg + m_ice)(-234221)
= -70266.3 - 234221m_ice

And then add this Q₃ to the others and set =0?

Be careful here.

The warm water in the beaker is being cooled from 78.6 °C to 22.7 °C.
The melt water is being warmed from 0 °C to 22.7 °C.

Your equations should reflect this.

 

[madison222s]

Be careful here.

The warm water in the beaker is being cooled from 78.6 °C to 22.7 °C.
The melt water is being warmed from 0 °C to 22.7 °C.

Your equations should reflect this.

So what I really need is 2 extra equations then?

 

[madison222s]

So what I really need is 2 extra equations then?

No, nevermind. Just one. For the ice water. Q₄= m_icec_ice(22.7-0)

 

[SteamKing]

No, nevermind. Just one. For the ice water. Q₄= m_icec_ice(22.7-0)

Remember, the ice has already melted. c_ice should be c_water.

 

[madison222s]

Remember, the ice has already melted. c_ice should be c_water.

Oh of course. Ok I got it, thanks!