Final answer:
Pouring 0.0100 kg of 20.0°C water onto a 1.20-kg block of ice will result in the water cooling down to a temperature close to the initial temperature of the ice, likely between -15.0°C and 0°C, as the water does not have enough energy to warm the ice to its melting point.
Step-by-step explanation:
If you pour 0.0100 kg of water at 20.0°C onto a 1.20-kg block of ice initially at -15.0°C, and effects of the surroundings are negligible, you are essentially asking how the temperature of the system changes. To solve this, we need to use the concept of thermal equilibrium where the system will reach a balance of temperatures, and also consider the heat capacity of water and ice, and the latent heat of fusion for ice.
To solve this problem, we assume that all the thermal energy lost by the water will be gained by the ice. The water will cool down, and the ice will warm up until they reach the same temperature. This process will continue until the ice reaches 0°C, at which point, if there is any remaining thermal energy in the water, it will begin to melt the ice.
However, since the mass of water being poured onto the ice is relatively small compared to the mass of the ice, it's highly unlikely that any melting will occur. Therefore, the final temperature will be somewhere between -15.0°C and 0°C, likely closer to -15.0°C due to the mass and initial temperature difference between the ice and the water.
The specific heat of ice is approximately 2.09 J/(g°C), so the energy required to raise the temperature of 1.20 kg of ice from -15.0°C to 0°C would be:
1.20 kg * 2.09 J/(g°C) * 15°C = 37,620 J
The energy given off by cooling 0.0100 kg of water from 20.0°C to 0°C would be:
0.0100 kg * 4.18 J/(g°C) * 20°C = 836 J
This clearly shows that the water does not have enough energy to warm the ice to 0°C, and therefore, no phase change will occur. The water will just reduce its temperature to the final temperature of the ice. Calculating the exact final temperature would require setting up a heat balance equation and solving for the final temperature.