How much ice (at 0°C) must be added to 1.15 kg of water at 81 °C so as to end up with all liquid at 26 °C? (ci = 2000 J/(kg.°C), cw = 4186 J/(kg.°C), Lf=3.35×10^5 J/kg, Lv= 2.26×10^6 J/kg)
The sum of the heats gained will be zero (some will be negative gained).
heaticemeleting+heat waterwarmcooling+heaticewaterwarming=0
M*3.35e5+1.15*4186*(26-81)+M*4186(26-0)=0
M(3.35e5+4186*26)=1.15*4186(55)
M= ... ? I get about 600 grams (.6kg)
To solve this problem, we need to determine the amount of heat lost by the hot water, the amount of heat gained by the ice, and then calculate the mass of ice required to achieve the final temperature.
Let's break down the steps:
Step 1: Calculate the heat lost by the hot water.
The formula to calculate heat transfer is:
Q = m * c * ΔT
Where:
Q is the heat transfer
m is the mass
c is the specific heat capacity
ΔT is the change in temperature
For the hot water, we have:
m1 = 1.15 kg (mass of water)
c1 = 4186 J/(kg.°C) (specific heat capacity of water)
ΔT1 = (81°C - 26°C) = 55°C (change in temperature)
Therefore, the heat lost by the hot water is:
Q1 = m1 * c1 * ΔT1
Step 2: Calculate the heat gained by the ice.
The formula to calculate heat transfer is the same as in Step 1:
Q = m * c * ΔT
For the ice, we have:
m2 = ? (mass of ice, what we're looking for)
c2 = 2.26×10^6 J/kg (specific heat capacity of ice)
ΔT2 = (26°C - 0°C) = 26°C (change in temperature)
Therefore, the heat gained by the ice is:
Q2 = m2 * c2 * ΔT2
Step 3: Calculate the heat gained by the ice due to the phase change.
The formula for heat transfer during a phase change is:
Q = m * L
Where:
Q is the heat transfer
m is the mass
L is the specific latent heat
For the ice, we have:
m2 = ? (mass of ice, what we're looking for)
Lf = 3.35×10^5 J/kg (specific latent heat of fusion)
Therefore, the heat gained by the ice during the phase change is:
Q3 = m2 * Lf
Now, let's set up the equation to solve for the mass of ice (m2):
Q1 + Q2 + Q3 = 0
m1 * c1 * ΔT1 + m2 * c2 * ΔT2 + m2 * Lf = 0
Substituting the given values:
1.15 kg * 4186 J/(kg.°C) * 55°C + m2 * (2.26×10^6 J/kg) * 26°C + m2 * (3.35×10^5 J/kg) = 0
Now we can solve for m2 (mass of ice).