A Styrofoam bucket of negligible mass contains 1.60 kg of water and 0.425 kg of ice. More ice, from a refrigerator at -14.8 degree C, is added to the mixture in the bucket, and when thermal equilibrium has been reached, the total mass of ice in the bucket is 0.768 kg.
Assuming no heat exchange with the surroundings, what mass of ice was added?
m=??kg
To find the mass of ice that was added, we need to use the concept of heat transfer and thermal equilibrium.
The heat gained by the additional ice (which was initially at -14.8 degrees Celsius) is equal to the heat lost by the water and the initial ice present in the bucket.
The heat gained or lost can be calculated using the equation:
Q = m * c * ΔT
Where:
Q = heat gained or lost
m = mass
c = specific heat capacity
ΔT = change in temperature
In this case, the heat gained by the ice will be equal to the heat lost by the water and initial ice. Since there is no heat exchange with the surroundings, the total heat gained and lost should be equal.
For the water and initial ice:
Q_water + Q_initial_ice = 0
And for the additional ice:
Q_additional_ice = 0
The equation for the heat gained by the additional ice can be written as:
Q_additional_ice = m_additional_ice * c_ice * ΔT_ice
Since the ice is in thermal equilibrium, its temperature is the same as the final temperature of the water and initial ice.
Using the given values, we can calculate the heat gained by the water and initial ice:
Q_water + Q_initial_ice = (m_water * c_water * ΔT_water) + (m_initial_ice * c_ice * ΔT_ice)
Since the temperature change (ΔT_water and ΔT_ice) is the same for both the water and the initial ice:
(m_water * c_water * ΔT) + (m_initial_ice * c_ice * ΔT) = 0
We can rearrange this equation to solve for the mass of the additional ice (m_additional_ice):
m_additional_ice = -(m_water * c_water * ΔT) / (c_ice * ΔT)
Now we can substitute the given values into the equation to find m_additional_ice:
m_water = 1.60 kg (mass of water)
m_initial_ice = 0.425 kg (mass of initial ice)
c_water = 4186 J/kg°C (specific heat capacity of water)
c_ice = 2100 J/kg°C (specific heat capacity of ice)
ΔT = final temperature - initial temperature
ΔT = 0°C – (-14.8°C) = 14.8°C
m_additional_ice = - (1.60 kg * 4186 J/kg°C * 14.8°C) / (2100 J/kg°C * 14.8°C)
By cancelling out the units and performing the calculation, we get:
m_additional_ice = - (52268.8 J) / (31080 J)
m_additional_ice ≈ -1.679 kg
Since mass cannot be negative, we discard the negative sign in the result:
m_additional_ice ≈ 1.679 kg
Therefore, the mass of ice that was added to the mixture in the bucket is approximately 1.679 kg.