A cube of ice is taken from the freezer at -7.5°C and placed in a 101 g aluminum calorimeter filled with 282 g of water at room temperature of 20°C. The final situation is observed to be all water at 16°C. What was the mass of the ice cube?
The sum of heats gained is zero.
let m be the mass of the ice cube
m*specificheatice*(7.5)+m*Lfice+101*spcificheatAl*(16-20)+282specificheatwater*(16-20)=0
check that.
To solve this problem, we can use the principle of conservation of energy.
First, we need to calculate the heat lost by the ice cube by using the formula:
Q1 = m1 * c1 * ΔT1
Where:
Q1 = heat lost by the ice cube
m1 = mass of the ice cube
c1 = specific heat capacity of ice
ΔT1 = change in temperature of the ice cube
The specific heat capacity of ice is 2.09 J/g°C.
The change in temperature of the ice cube can be calculated as:
ΔT1 = (16°C - (-7.5°C)) = 23.5°C
So the equation becomes:
Q1 = m1 * 2.09 J/g°C * 23.5°C
Next, we need to calculate the heat gained by the water in the calorimeter by using the formula:
Q2 = m2 * c2 * ΔT2
Where:
Q2 = heat gained by the water
m2 = mass of the water
c2 = specific heat capacity of water
ΔT2 = change in temperature of the water
The specific heat capacity of water is 4.18 J/g°C.
The change in temperature of the water can be calculated as:
ΔT2 = (16°C - 20°C) = -4°C
Since the water lost heat in this case, we need to use the absolute value of ΔT2.
So the equation becomes:
Q2 = m2 * 4.18 J/g°C * |-4°C|
According to the principle of conservation of energy, the heat lost by the ice cube is equal to the heat gained by the water:
Q1 = Q2
m1 * 2.09 J/g°C * 23.5°C = m2 * 4.18 J/g°C * |-4°C|
Simplifying the equation:
m1 * 2.09 * 23.5 = m2 * 4.18 * 4
Now, we can substitute the known values:
m1 * 2.09 * 23.5 = 282 g * 4.18 * 4
Now, we solve for m1:
m1 = (282 g * 4.18 * 4) / (2.09 * 23.5)
m1 ≈ 280 g
So, the mass of the ice cube is approximately 280 grams.
To solve this problem, we need to use the principle of energy conservation. The energy lost by the ice as it warms up and melts is equal to the energy gained by the water and the calorimeter.
Here's how to approach this problem step by step:
Step 1: Calculate the energy gained by the water and the calorimeter.
The energy gained by the water and the calorimeter can be calculated using the formula:
Q = mcΔT
Where:
Q is the energy gained (or lost)
m is the mass of the substance (water or calorimeter)
c is the specific heat capacity of the substance
ΔT is the change in temperature
For water, c = 4.184 J/g°C
For the calorimeter (aluminum), c = 0.897 J/g°C
Calculate the energy gained by the water using the following formula:
Q_water = m_water * c_water * ΔT_water
Where:
m_water is the mass of the water (282 g)
c_water is the specific heat capacity of water (4.184 J/g°C)
ΔT_water is the change in temperature of the water (16 - 20 = -4°C)
Q_water = 282 g * 4.184 J/g°C * (-4°C) = -4714.944 J (Note: The negative sign indicates energy loss)
Calculate the energy gained by the calorimeter using the following formula:
Q_calorimeter = m_calorimeter * c_calorimeter * ΔT_calorimeter
Where:
m_calorimeter is the mass of the calorimeter (101 g)
c_calorimeter is the specific heat capacity of aluminum (0.897 J/g°C)
ΔT_calorimeter is the change in temperature of the calorimeter (16 - 20 = -4°C)
Q_calorimeter = 101 g * 0.897 J/g°C * (-4°C) = -363.788 J (Note: The negative sign indicates energy loss)
Step 2: Calculate the energy lost by the ice.
The energy lost by the ice can be calculated using the formula:
Q_ice = m_ice * ΔH_f + m_ice * c_ice * ΔT_ice
Where:
m_ice is the mass of the ice (unknown, what we're trying to find)
ΔH_f is the heat of fusion for ice (334 J/g)
c_ice is the specific heat capacity of ice (2.093 J/g°C)
ΔT_ice is the change in temperature of the ice (-7.5 - 0 = -7.5°C)
Q_ice = m_ice * 334 J/g + m_ice * 2.093 J/g°C * (-7.5°C)
Q_ice = 334 m_ice - 15.6975 m_ice = 318.3025 m_ice
Step 3: Apply the principle of energy conservation.
According to the principle of energy conservation, the energy lost by the ice (Q_ice) is equal to the energy gained by the water (Q_water) and the calorimeter (Q_calorimeter).
Q_ice = Q_water + Q_calorimeter
318.3025 m_ice = -4714.944 J + (-363.788 J)
318.3025 m_ice = -5078.732 J
m_ice = -5078.732 J / 318.3025
m_ice = -15.9434 g
Since mass cannot be negative, we can ignore the negative sign.
Therefore, the mass of the ice cube is approximately 15.9434 g.