An insulated beaker with negligible mass contains
0.250 kg of water at a temperatur 75.0°C.e of How many kilograms
of ice at a temperature of -20.0°C must be dropped into the water
to make the final temperature of the system 40.0°C?
X kilograms of ice
SH means specific heat or heat capacity
HF means heat of fusion
Heat into ice
= X(20)SHice + X(HFice) + X(40)SHwater
=
Heat out of water
= 0.250(SHwater)(75-40)
To solve this problem, we can use the principle of conservation of energy. The energy gained by the ice must be equal to the energy lost by the water.
Let's calculate the energy gained by the ice:
Energy gained by the ice = mass of ice × specific heat capacity of ice × change in temperature
Given:
Mass of water = 0.250 kg
Initial temperature of water = 75.0°C
Final temperature of system = 40.0°C
Specific heat capacity of ice = 2.09 kJ/kg·°C (from reference)
To find the mass of ice, we need to calculate the energy lost by the water:
Energy lost by the water = mass of water × specific heat capacity of water × change in temperature
Specific heat capacity of water = 4.18 kJ/kg·°C (from reference)
The energy lost by the water must be equal to the energy gained by the ice:
Energy lost by the water = Energy gained by the ice
Mass of water × specific heat capacity of water × change in temperature = Mass of ice × specific heat capacity of ice × change in temperature
Rearranging the equation:
Mass of ice = (Mass of water × specific heat capacity of water × change in temperature) / (specific heat capacity of ice × change in temperature)
Now let's substitute the given values into the equation and calculate the mass of ice:
Mass of ice = (0.250 kg × 4.18 kJ/kg·°C × (75.0°C - 40.0°C)) / (2.09 kJ/kg·°C × (40.0°C - (-20.0°C)))
Mass of ice = (0.250 kg × 4.18 kJ/kg·°C × 35.0°C) / (2.09 kJ/kg·°C × 60.0°C)
Mass of ice = 0.104 kg
Therefore, approximately 0.104 kg of ice at a temperature of -20.0°C must be dropped into the water to make the final temperature of the system 40.0°C.
To solve this problem, we need to use the principle of conservation of energy. The heat gained by the ice will be equal to the heat lost by the water.
First, we need to calculate the heat lost by the water. We can use the equation:
Q = mcΔT
Where:
Q is the heat lost (or gained),
m is the mass of the substance,
c is the specific heat capacity of the substance, and
ΔT is the change in temperature of the substance.
For water, the specific heat capacity (c) is about 4.18 J/g°C.
Given:
Mass of water (m1) = 0.250 kg
Initial temperature of water (T1) = 75.0°C
Final temperature of the system (Tf) = 40.0°C
Using the equation, we can calculate the heat lost by the water:
Q1 = m1 * c * ΔT1
= 0.250 kg * 4.18 J/g°C * (40.0°C - 75.0°C)
Note that we converted the specific heat capacity from J/g°C to J/kg°C by multiplying it by 1000 because the mass is given in kilograms.
Now, we need to calculate the heat gained by the ice. We can use the equation:
Q2 = m2 * Lf
Where:
Q2 is the heat gained (or lost),
m2 is the mass of the substance (in this case, ice), and
Lf is the latent heat of fusion of the substance.
The latent heat of fusion (Lf) for water is about 334,000 J/kg.
Since the ice is at -20.0°C and we want to raise its temperature to 40.0°C, the change in temperature (ΔT2) is 40.0°C - (-20.0°C).
Now, we can solve for the mass of ice (m2) using the equation:
Q1 = Q2
m1 * c * ΔT1 = m2 * Lf
Rearranging the equation:
m2 = (m1 * c * ΔT1) / Lf
Substituting the given values:
m2 = (0.250 kg * 4.18 J/g°C * (40.0°C - 75.0°C)) / 334,000 J/kg
By solving the equation, you will find the mass of ice required to make the final temperature of the system 40.0°C.