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A thermally insulated vessel containing 100 grams of ice at 0 degrees Celsius . Water at 55 degrees celsius is poured and after some time , the system reaches the equilibrium at a temperature of 10 degrees Celsius. That amount of water in grams was placed . ( heat fusion ice 80cal / g . The specific heat capacity of water is about 1cal / ( gr.K )
Here suppose mass of ice formed=m gram
Therefore, mass of water evaporated= 100-m gram
As, no water is left in he vessel ,
Amt. Of heat taken by water in evaporation = amt. Of heat given out by water in freezing
(100-m)×2.1×10^6/10^3=m×3.36×10000/1000
m=2100/24.36
m=86.2g
Bad
To find the amount of water that was placed in the thermally insulated vessel, we can use the principle of heat transfer.
First, let's calculate the heat gained by the ice to reach 10 degrees Celsius:
Heat gained by the ice = mass of ice x heat fusion of ice
Given:
Mass of ice = 100 grams
Heat fusion of ice = 80 cal/g
Heat gained by the ice = 100 grams x 80 cal/g
Heat gained by the ice = 8000 calories
Next, let's calculate the heat lost by the hot water to reach 10 degrees Celsius:
Heat lost by the hot water = mass of water x specific heat capacity of water x change in temperature
Given:
Initial temperature of water = 55 degrees Celsius
Final temperature of the system = 10 degrees Celsius
Specific heat capacity of water = 1 cal/(g.K)
Change in temperature = Final temperature - Initial temperature
Change in temperature = 10 degrees Celsius - 0 degrees Celsius
Change in temperature = 10 degrees Celsius
Heat lost by the hot water = mass of water x 1 cal/(g.K) x 10 degrees Celsius
Heat lost by the hot water = 10 x mass of water calories
Since the system reaches equilibrium, the heat gained by the ice is equal to the heat lost by the hot water:
8000 calories = 10 x mass of water calories
Now, let's solve for the mass of water:
mass of water = 8000 calories / 10 calories/gram
mass of water = 800 grams
Therefore, 800 grams of water was placed in the thermally insulated vessel.
To find the amount of water in grams that was placed, we need to use the principle of energy conservation. In this situation, the heat lost by the hot water will be equal to the heat gained by the ice and the cold water.
Let's break down the steps to find the solution:
Step 1: Calculate the heat lost by the hot water.
- The initial temperature of the hot water is 55 degrees Celsius.
- The final temperature of the system (after reaching equilibrium) is 10 degrees Celsius.
- The specific heat capacity of water is 1 cal/(g.K).
Using the formula Q = m * c * ΔT, where Q is the heat, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature, we can calculate the heat lost by the hot water.
Q_hot = m_hot * c_water * ΔT_hot
ΔT_hot = 55 - 10 = 45 degrees Celsius
Step 2: Calculate the heat gained by the ice and the cold water.
- The final temperature of the system (after reaching equilibrium) is 10 degrees Celsius.
- The specific heat capacity of water is 1 cal/(g.K).
- The heat of fusion (melting) for ice is 80 cal/g.
For the ice, the heat gained is due to both the temperature change (from 0 to 10 degrees Celsius) and the phase change (from solid to liquid). We need to include both contributions.
Q_ice = Q_temp_change + Q_phase_change
Q_temp_change = m_ice * c_water * ΔT_ice
Q_phase_change = m_ice * heat_fusion_ice
ΔT_ice = 10 - 0 = 10 degrees Celsius
Step 3: Equate the heat lost and heat gained to find the mass of water.
Q_hot = Q_ice + Q_water
m_hot * c_water * ΔT_hot = (m_ice * c_water * ΔT_ice) + (m_water * c_water * ΔT_water)
Since specific heat capacity and temperature changes are the same for water, we can simplify the equation to:
m_hot * ΔT_hot = (m_ice * ΔT_ice) + (m_water * ΔT_water)
Now, we can substitute the known values and solve for m_water.
m_hot * ΔT_hot = (m_ice * ΔT_ice) + (m_water * ΔT_water)
m_water = (m_hot * ΔT_hot - m_ice * ΔT_ice) / ΔT_water
Substituting the values:
m_water = (100g * 45°C - 100g * 10°C) / (10°C - 55°C)
m_water = -3500g / -45°C
m_water = 77.78g
Therefore, the amount of water in grams that was placed is approximately 77.78 grams.