Steam at 100 degrees C is passed into a vessel containing 1 kilogram of ice at 0 degrees until the ice is just melted. Find the weight of the water formed.

Any idea what equation I would use to find this?

You need:

Q = mL

Where Q is the energy required for the steam (or ice) to melt, m the mass of the steam (or ice) and L is the specific latent heat of the substance (that of ice is different from that of steam)

I guess your question is asking the total weight of water, that is from steam and ice that is collected when the ice melts.

Well, using Q_{steam} = m_{steam}L_{steam} and Q_{ice} = m_{ice}L_{ice}, you will see that Q_steam and Q_ice should be equal, because the energy carried by the steam is transferred to the ice, hence the energy lost by steam must be equal to energy lost by ice.

So, you get:

m_{steam}L_{steam} = m_{ice}L_{ice}

L_steam and L_ice should be given to you, and you know m_ice, find m_steam. The total weight of water is given by (m_ice + m_steam)g.

Post your answer! :)

Thank you!

This is what I got:
QSteam=(.148Kg)(2260J)=2260J
QIce=(1Kg)(335J)=335J

So .148Kg+1Kg=1.48
W=1.48(9.8)=14.5

Correct, you found the mass to be 0.148 kg, which is good. But you didn't add them up correcly. 0.148 kg + 1 kg = 1.148 kg. Which then has the weight of 11 kg (2 sig fig).

Be careful next time :)