An insulated container has 54 grams of 90 ªC water. How much ice is required to lower the temperature of 50ºC? No ice remains.
To determine the amount of ice required to lower the temperature of 90 ºC water to 50 ºC, we need to use the principle of heat transfer.
The formula we will use is:
Q = mcΔT
Where:
Q = heat transferred
m = mass of the substance (water or ice)
c = specific heat capacity of the substance
ΔT = change in temperature
First, let's calculate the heat lost by the water when it cools down from 90 ºC to 50 ºC.
Q_water = m_water * c_water * ΔT_water
Given:
m_water = 54 grams
c_water = 4.18 J/gºC (specific heat capacity of water)
ΔT_water = 90 ºC - 50 ºC = 40 ºC
Plugging the values into the formula:
Q_water = 54 g * 4.18 J/gºC * 40 ºC
Q_water = 8985.6 J
Now, since the heat loss by the water will be gained by the ice to melt it, we can calculate the mass of ice required to absorb this amount of heat.
Q_ice = m_ice * ΔH_fusion
Where:
Q_ice = heat gained by the ice
m_ice = mass of ice
ΔH_fusion = heat of fusion (amount of heat required to melt 1 gram of ice)
According to the given information, all the ice will melt, so Q_ice = Q_water.
m_ice * ΔH_fusion = 8985.6 J
Now, we need to find the heat of fusion for ice. The heat of fusion for ice is 333.5 J/g.
m_ice * 333.5 J/g = 8985.6 J
Simplifying the equation, we find:
m_ice = 8985.6 J / 333.5 J/g
m_ice ≈ 26.96 g
So approximately 26.96 grams of ice are required to lower the temperature of 90 ºC water to 50 ºC.