- #1

bluesteels

- 28

- 1

- Homework Statement
- A copper calorimeter can with mass 0.100 kg contains 0.160 kg of water and 0.0180 kg of ice in thermal equilibrium at atmospheric pressure. If 0.750 kg of lead at 255°C is dropped into the calorimeter can, what is the final temperature? Assume that no heat is lost to the surroundings?"

- Relevant Equations
- Q=mc delta T

Q=mL (latent heat)

My thought process of how i do the ice melting part: (note I just ignore the copper/lead part cause I already know how to do that part)

Q_ice + Q_melt + Q_liquid so, it 0.018(

but on chegg they didn't use

Like i know it because ice melting into water but if that the case how come this problem below they don't even use it

"In a container of negligible mass, 0.200 kg of ice at an initial temperature of -40.0°C is mixed with a mass m of water that has an initial temperature of 80.0°C. No heat is lost to the surroundings. If the final temperature of the system is 28.0°C, what is the mass m of the water that was initially at 80.0°C?"

For this problem they did

0.2(

Q_ice + Q_melt + Q_liquid so, it 0.018(

**2100**)T+0.16(4190)T+0.018(334*10^3)but on chegg they didn't use

**2100**but they just use 4190 instead and I am confused on why they did that.Like i know it because ice melting into water but if that the case how come this problem below they don't even use it

"In a container of negligible mass, 0.200 kg of ice at an initial temperature of -40.0°C is mixed with a mass m of water that has an initial temperature of 80.0°C. No heat is lost to the surroundings. If the final temperature of the system is 28.0°C, what is the mass m of the water that was initially at 80.0°C?"

For this problem they did

0.2(

**2100**)(40)+0.2(334*10^3)+0.2(4190)(80). if that was the case earlier then wouldn't you sub. 2100 for 4190 for this problem.