calculate the heat of fusion of ice from the following data at 0°c add to water :
Mass of calorimetet 60 g
Mass of calorimeter and water 460
Mass of calorimeter plus water and ice 618 g
Initial temperature of water 38 °c
Final temperature of mixture 5 °c
Specific heat of calorimeter 0.10 cal/g.°c
80 cal/g
first begin by identify the mass of every given substance.
mass calorimeter = 60g
mass water = 460g-60g= 400g
mass ice= 618g-460g=158g
Qgain=Qlost
Qice=Qwater
therefore
mL + m(detlta temp)C=(delta temp)(mass water * C +mass calor *C)
and then we make L subject of the formula
so
L=[33(0.4*4186 + 0.06*400)-(0.158*5*4186)] / 0.158
L=333799.1139 J/kg
To calculate the heat of fusion of ice, we need to use the principle of energy conservation. The heat lost by the water as it cools to its final temperature is gained by the ice as it melts.
Step 1: Calculate the heat lost by the water.
Q_water = m_water * c_water * ΔT_water
where
m_water = mass of water
c_water = specific heat capacity of water
ΔT_water = change in temperature of water
We need to find the mass of water, which is the mass of the calorimeter and water minus the mass of the calorimeter.
m_water = (mass of calorimeter and water) - (mass of calorimeter)
m_water = 460 g - 60 g
m_water = 400 g
Now we can calculate the heat lost by the water:
Q_water = 400 g * 1 cal/g.°c * (5 °c - 38 °c)
Q_water = 400 g * 1 cal/g.°c * (-33 °c)
Q_water = -13,200 cal
Step 2: Calculate the heat gained by the ice.
Q_ice = m_ice * L_fusion
where
m_ice = mass of ice
L_fusion = latent heat of fusion of ice
We need to find the mass of ice, which is the mass of the calorimeter, water, and ice minus the mass of the calorimeter and water.
m_ice = (mass of calorimeter, water, and ice) - (mass of calorimeter and water)
m_ice = 618 g - 460 g
m_ice = 158 g
To calculate the heat gained by the ice, we need to use the heat lost by the water (Q_water) from Step 1:
Q_ice = -Q_water
Q_ice = -(-13,200 cal)
Q_ice = 13,200 cal
Now, we can substitute the values into the equation and solve for L_fusion:
13,200 cal = 158 g * L_fusion
L_fusion = 13,200 cal / 158 g
L_fusion ≈ 83.54 cal/g
Therefore, the heat of fusion of ice is approximately 83.54 calories per gram.
To calculate the heat of fusion of ice, we can use the equation:
Heat (Q) = m * c * ΔT
where Q is heat, m is mass, c is specific heat, and ΔT is the change in temperature.
First, let's calculate the heat absorbed by the water:
Mass of water = Mass of calorimeter and water - Mass of calorimeter
= 460 g - 60 g
= 400 g
ΔT of water = Final temperature - Initial temperature
= 5 °C - 38 °C
= -33 °C
Now, we can calculate the heat absorbed by the water using the equation:
Qwater = m * c * ΔT
= 400 g * 1 cal/g.°C * (-33 °C)
= -13,200 cal
Since ΔT is negative, it means the water lost heat.
Next, let's calculate the heat released by the ice:
Mass of ice = Mass of calorimeter plus water and ice - Mass of calorimeter and water
= 618 g - 460 g
= 158 g
ΔT of ice = Final temperature - Initial temperature
= 5 °C - 0 °C
= 5 °C
The specific heat of ice is negligible, so we can assume it is 0.
Qice = m * c * ΔT
= 158 g * 0 cal/g.°C * 5 °C
= 0 cal
Now, let's calculate the heat lost by the calorimeter:
Qcalorimeter = Mass of calorimeter * specific heat * ΔT
= 60 g * 0.10 cal/g.°C * (-33 °C)
= -198 cal
Again, since ΔT is negative, it means the calorimeter lost heat.
Finally, to calculate the heat of fusion of ice, we can use the equation:
Heat of fusion (Qfusion) = -(Qwater + Qcalorimeter)
= -( -13,200 cal + -198 cal )
= 13,398 cal
Therefore, the heat of fusion of ice is 13,398 cal.