What mass of steam at \(100^{\circ} \mathrm{C}\) must be mixed with \(150 \mathrm{~g}\) of ice at \(0^{\circ} \mathrm{C}\), in a thermally insulated container, to produce liquid water at \(50^{\circ} \mathrm{C}\) ?

Short Answer

Expert verified
The result of the calculation in step 5 will yield the mass of the steam required. Note that as the calculation involves energy being conserved, the result is dependent on the specific values of the system.

Step by step solution

01

Define the total energy exchange

The total energy exchange can be defined as the sum of the energy used to melt the ice, heat the water from \(0^{\circ}C\) to \(50^{\circ}C\), and condense the steam to water then cool it to \(50^{\circ}C\). Thus, the equation can be written as: \[Q_{total}=Q_{ice->water} + Q_{water-heating} + Q_{steam->water} + Q_{cooling-steam}\]
02

Calculate the energy to melt the ice

The energy required to convert ice at \(0^{\circ}C\) to water at \(0^{\circ}C\) can be calculated using the formula: \[Q_{ice->water} = m_{ice} * L_{f}\] where \(m_{ice} = 150g\) is the mass of the ice and \(L_{f} = 334 J/g\) is the latent heat of fusion of ice.
03

Calculate the energy to heat the water from \(0^{\circ}C\) to \(50^{\circ}C\)

The energy needed to heat the resulting water from \(0^{\circ}C\) to \(50^{\circ}C\) can be calculated using the formula: \[Q_{water-heating} = m_{water} * c_{water} * \Delta T\] where \(m_{water} = 150g\) is the mass of the resulting water, \(c_{water} = 4.18 J/g.\degree C\) is the specific heat capacity of water, and \(\Delta T = 50^{\circ}C - 0^{\circ}C = 50^{\circ}C\) is the change in temperature.
04

Calculate the energy to convert steam to water then cool it to \(50^{\circ}C\)

The energy taken by the steam to condense to water at \(100^{\circ}C\) and then cool down to \(50^{\circ}C\) is calculated with the formula: \[Q_{steam->water} = m_{steam} * L_v\] and \[Q_{cooling-steam} = m_{steam} * c_{water} * \Delta T\] respectively. Here, \(m_{steam}\) is the mass of steam we are asked to find, \(L_v = 2260 J/g\) is the latent heat of vaporization of water, \(c_{water}\) is the specific heat capacity of water, and \(\Delta T = 100^{\circ}C - 50^{\circ}C = 50^{\circ}C\) is the change in temperature.
05

Solve for the mass of the steam

Setting the energy given by the steam to be equal to the energy taken in by the ice and the resulting water, the mass of the steam, \(m_{steam}\), is found by solving the equation: \[Q_{total}=0\] This will lead to the value of the mass of the steam needed.

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