0.115 kg of water at 89.0o C is poured into an insulated cup containing 0.226 kg of ice initally at 0o. Calculate the mass of liquid when the system reaches thermal equilibrium.
sum of heats gained is zero.
MassIceMelted*Hf+masswater*specificheat*changtemp=0
first assume to see ifthe mass of ice melted is less than what you started with.
Mmelt*333.55 kJ/kg+0.115*4.18kJ/kg*(0-89)=0
massicemelted= .115*4.18*89/(333.55)=.128kg, which is only part of the ice you started with.
the mass of liquid is food
Thanks anonymous and bobpurspley
To solve this problem, we need to consider the heat gained and lost by each component of the system and set them equal to each other once thermal equilibrium is reached. The heat gained/lost can be calculated using the specific heat capacity and the equation:
Q = m * c * ΔT
Where:
Q is the heat gained or lost
m is the mass of the substance
c is the specific heat capacity
ΔT is the change in temperature
Let's first calculate the heat gained by the water. We know the mass of the water is 0.115 kg, the specific heat capacity of water is 4190 J/kg°C, and the change in temperature is:
ΔT = 89.0°C - 0°C = 89.0°C
Q_water = m_water * c_water * ΔT
Q_water = 0.115 kg * 4190 J/kg°C * 89.0°C
Q_water = 41605.15 J
Next, we calculate the heat lost by the ice. We know the mass of the ice is 0.226 kg, the specific heat capacity of ice is 2100 J/kg°C, and the change in temperature is:
ΔT = 0°C - 0°C = 0°C
Q_ice = m_ice * c_ice * ΔT
Q_ice = 0.226 kg * 2100 J/kg°C * 0°C
Q_ice = 0 J
Since no heat is lost by the ice, we can set Q_water equal to Q_ice:
Q_water = Q_ice
41605.15 J = 0 J
At thermal equilibrium, all the water has turned into ice, which means the mass of the liquid is zero (since it has all solidified).