Final Answer:
The mass of steam to be added to 1 kg of ice at 0 degrees Celsius to yield liquid water at 30 degrees Celsius is 1.14 kg.
Step-by-step explanation:
In order to answer this question, the physical principle of energy conservation has to be applied. This principle states that the energy of a system before a collision is equal to the energy of the system after the collision. In this problem, the system is made up of the ice and steam, which are colliding on a macroscopic level.
The initial energy of the system is the energy of the ice at 0°C. The energy of the ice is calculated using the equation E = m*c*ΔT, where m is the mass of the ice, c is the specific heat capacity of water, and ΔT is the change in temperature. The energy of the ice is 0.33 kJ.
The final energy of the system is the energy of the liquid water at 30°C. The energy of the liquid water is calculated using the equation E = m*c*ΔT, where m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature. The energy of the liquid water is 4.58 kJ.
To find the mass of the steam, the equation mass = energy/heat of vaporization is used. The heat of vaporization of water is 2256 kJ/kg. Using this equation, the mass of the steam is 1.14 kg. This is the mass of steam that must be added to 1 kg of ice at 0 degrees Celsius to yield liquid water at 30 degrees Celsius.