You take 2 liters of water at 20 degrees Celsius and remove heat at 200W from it for 5 min. What is the temperature and state of the water after the experiment? Some properties of water: Heat of fusion: 333.55 kJ/kg, Heat of vaporization: 2257 kJ/kg, Heat capacity of steam: 2.080 kJ/kg, of water: 4.1813 kJ/kg, of ice: 2.05 kJ/kg.
To determine the final temperature and state of the water, we need to consider the changes in heat energy. Let's break down the steps involved in solving this problem:
Step 1: Calculate the heat removed from the water:
We know that the heat removed (Q) from an object is given by the formula:
Q = Power (P) x Time (t)
In this case, the power is given as 200W, and the time is 5 minutes (5 minutes = 300 seconds). Let's calculate Q by substituting these values:
Q = 200W x 300s = 60,000 Joules (J)
Step 2: Calculate the heat required to change the temperature of water:
The heat required (Q) to change the temperature of a substance can be calculated by using the formula:
Q = mass (m) x specific heat capacity (C) x temperature change (ΔT)
In this case, the mass of the water is 2 liters, which is equivalent to 2000 grams (1 liter of water = 1000 grams). The specific heat capacity of water is 4.1813 kJ/kg, which is equivalent to 4.1813 J/g. The initial temperature (T1) is 20 degrees Celsius, and the final temperature (T2) is what we need to determine. We can rearrange the formula to solve for T2:
Q = m x C x ΔT
ΔT = T2 - T1
T2 = ΔT + T1
Step 3: Determine the change in temperature:
To find the change in temperature (ΔT), we can use the equation:
ΔT = Q / (m x C)
Substituting the values, we get:
ΔT = 60,000 J / (2000 g x 4.1813 J/g)
ΔT ≈ 7.18 degrees Celsius
Step 4: Calculate the final temperature:
Now that we have the change in temperature, we can determine the final temperature (T2):
T2 = ΔT + T1
T2 = 7.18 + 20
T2 ≈ 27.18 degrees Celsius
Step 5: Determine the state of the water:
To determine the state of the water, we need to consider the final temperature. At temperatures below 0 degrees Celsius, water exists as ice. At temperatures above 100 degrees Celsius, water exists as steam. Therefore, if the final temperature is below 0 degrees Celsius, the water is in the solid state (ice). If the final temperature is above 100 degrees Celsius, the water is in the gaseous state (steam). Otherwise, it remains in the liquid state.
In this case, the final temperature of the water is approximately 27.18 degrees Celsius, which falls within the range of 0 to 100 degrees Celsius. Therefore, the water remains in the liquid state after the experiment.