What mass of steam at 100°C must be added to 1.00 kg of ice at 0°C to yield liquid water at 15°C?

Question

What mass of steam at 100°C must be added to 1.00 kg of ice at 0°C to yield liquid water at 15°C?

Answer 1

The sum of heats gained is zero (on has negative heat gained).

Add the heats.

1kg*Lfice+1kg(cwater)(15-0)+M*Lvsteam+M*cwater*(15-100)=0
solve for M

Answer 2

To solve this problem, we will use the energy conservation equation:

Qice + Qsteam + Qwater = 0

Where:
Qice is the heat gained by the ice
Qsteam is the heat gained by the steam
Qwater is the heat gained by the water

First, let's calculate Qice:

Qice = m * specific_heat_ice * ΔTice

Where:
m is the mass of the ice
specific_heat_ice is the specific heat capacity of ice
ΔTice is the change in temperature of the ice

For ice, specific_heat_ice = 2.09 J/g · °C
ΔTice = 15°C - 0°C = 15°C

Now, let's calculate Qwater:

Qwater = m * specific_heat_water * ΔTwater

Where:
specific_heat_water is the specific heat capacity of water
ΔTwater = 15°C - 0°C = 15°C

For water, specific_heat_water = 4.18 J/g · °C

Now, let's calculate Qsteam:

Qsteam = m * specific_heat_steam * ΔTsteam

Where:
specific_heat_steam is the specific heat capacity of steam
ΔTsteam = 100°C - 15°C = 85°C

For steam, specific_heat_steam = 2.03 J/g · °C

Since Qsteam = -Qice - Qwater, we can solve for the mass of the steam:

m = (Qwater + Qice) / specific_heat_steam / ΔTsteam

Now, let's substitute the values into the equations:

Qice = m * 2.09 J/g · °C * 15°C
Qwater = m * 4.18 J/g · °C * 15°C
Qsteam = -Qice - Qwater

Substituting these equations into m = (Qwater + Qice) / specific_heat_steam / ΔTsteam, we can solve for the mass of the steam.

Answer 3

To determine the mass of steam needed to be added to the ice, we can follow these steps:

Step 1: Calculate the heat required to melt the ice (Q1)
The heat required to melt the ice can be calculated using the formula:
Q1 = m1 * ΔHf
where m1 is the mass of the ice and ΔHf is the heat of fusion of water (334 kJ/kg).

Given:
m1 = 1.00 kg

Q1 = (1.00 kg) * (334 kJ/kg) = 334 kJ

Step 2: Calculate the heat required to raise the temperature of the melted ice from 0°C to 15°C (Q2)
The heat required to raise the temperature of an object can be calculated using the formula:
Q2 = m2 * c * ΔT
where m2 is the mass of the water, c is the specific heat capacity of water (4.18 kJ/kg°C), and ΔT is the temperature change.

Since we want to raise the temperature of the melted ice from 0°C to 15°C, ΔT = 15°C - 0°C = 15°C.

Given:
ΔT = 15°C
c = 4.18 kJ/kg°C

Let's say the mass of steam needed is m3. Then, after the ice melts, we have 1.00 kg of water. Therefore, m2 = m1 + m3.

Q2 = (m1 + m3) * c * ΔT
= (1.00 kg + m3) * (4.18 kJ/kg°C) * (15°C)
= 62.7 kJ + 62.7 m3 kJ

Step 3: Calculate the heat released by the steam when it condenses (Q3)
When the steam condenses, it releases the same amount of heat that was needed to melt the ice. Therefore, Q3 = -Q1 (negative sign to represent heat release).

Q3 = -334 kJ

Step 4: Calculate the net heat exchange (Q_net)
The total heat exchange in the process is given by the sum of the individual heat exchanges:
Q_net = Q1 + Q2 + Q3

0 = (334 kJ) + (62.7 kJ + 62.7 m3 kJ) + (-334 kJ)
0 = 62.7 kJ + 62.7 m3 kJ

To get the mass of steam needed (m3), we solve the equation for m3:
62.7 m3 kJ = -62.7 kJ
m3 = -62.7 kJ / 62.7 kJ
m3 = -1 kg

Since mass cannot be negative, this implies that no steam needs to be added to the ice to yield liquid water at 15°C. The ice will simply melt on its own due to the surrounding temperature.