Ice cube on a scale Lets put an ice cube on a fairly precise scale and take readings over a period of several hours. The ice cube is situated in a saucer so that the water is retained after melting. 
What will the recorded curve look like?
Suppose ambients conditions: 25°C, RH40% and 1013 mbar.
 A: The way your question is phrased, it looks like you are expecting the mass to change. In that case the only change will be a slight mass loss due to evaporation, but the rate of evaporation is a variable - dependant on room temperature, air pressure humidity and how still the air is above the sample. 
A: The mass of the melted water will be slightly lower than the ice. This is because the Latent Heat of Evaporation is higher than the Latent Heat of Fusion, so more water will be evaporated as it is turns into the liquid form and hence more mass will be lost.
So gradually, as the water melts, the rate at which the mass is lost increases and hence the graph will be decreasing with the slope increasing slightly as time passes.
It is obvious that I am assuming all the other parameters in the environment like pressure, temperature and humidity to remain constant.
A: The saucer ('cup') contains the ice at the beginning, which will melt so it contains water.
As we are interested in what the scale shows us - the change in mass - it is not important whether it's ice or not, the mass would not change.
The change we could observe is from evaporating water.
How quick the water evaporates depends on the surface area, the temperature. and the flow of air.
The surface area will decrease, as the ice melts - the surface becomes more flat, and smaller.
The average temperature of the surface will raise while the ice is melting. The water from molten ice will get warmer, while the surface of the ice stays at 0 degree C. As the surface of the ice gets smaller, contributing less to the average.
(This part is referring to a cup with higher border used in an earlier version of the question. It was later replaced by a saucer with a very low border, see below for the effect of the change:)
The evaporated water gets moved out of the cup as part of moist air moving out of it.
As long as the water is cold in relation to the environment, the air inside the cup will be cold, and thus of higher density, which causes only little water be moved out of the cup actually.
Related to that, there is the effect of the shape of the cup.
A higher cup causes slower evaporation by more keeping the moist, cold air from moving out.
A larger base area of the cup leads to a larger water/ice surface, causing faster evaporation.
To estimate the mass curve, all these effects need to be estimated, and - ignoring influences between them - added up in terms of evaporation per time.
(Update assuming the cup is replaced by a saucer that has a very low border:)
If we assume the border of the saucer is very low, just above the maximal water line (*) that would mean the evaporated water can leave the saucer pretty quickly. 
It would cause a change in speed of evaporation during the time at the start, when the ice is still present. With a high cup, the wet ice surface would be covered by cold, moist air. With the low border, there is convection moving the cooling air down along the ice surface, where it can transport any water vapor avay easily.
[(*) The water line is maximal when the remaining ice starts to swim in the water. Before, it increases as the ice cube adds water by melting, after that, the level get's lower by evaporation.]
A: If a cold beer can is taken out of the fridge on a warm day small water droplets form on its surface since the surface temperature is below the dew point of the moist ambient air. After a while, when the can reaches equilibrium temperature the water gets evaporated away again. 
For the ice cube the same is true in principle besides it evaporates away completely after some time.  
In the actual experiment (with T=25°C and RH=50%) the maximum mass was reached after about 50 mins and the mass gain was quite significant (+1%).

A: The matter is constant.  The mass increases (latent enthalpy of fusion plus specific heat as temperature rises) but ${m = E/c^2}$, hence way too small to measure.  What is big enough to measure is the deceasing net buoyancy for displaced air by the denser mass (lesser volume) of  of water versus ice, about 1.3 ${mg/cm^3}$ differential air volume at STP.  If the cup is sealed that dos not obtain.
