One liter of water at 47 degree C is used to make iced tea. How much ice at 0 degree C must be added to lower the temperature of the tea to 12 degree C? The specific heat of water is 1 cal/g · degree C and latent heat of ice is 79.7 cal/g.
To solve this problem, we need to use the principles of specific heat and latent heat.
First, let's determine the amount of heat that needs to be transferred from the water to lower its temperature from 47°C to 12°C. To do this, we can use the formula:
Q = mcΔT
where:
Q is the heat transferred
m is the mass of the substance
c is the specific heat of the substance
ΔT is the change in temperature
In this case, the water is being cooled down, so the heat transferred will be negative. Let's substitute the given values into the formula:
Q = -1g × 1 cal/g·°C × (12°C - 47°C) = -35 cal
Since the water is losing heat, the amount of ice we need to add must absorb this amount of heat. The heat needed to melt ice at 0°C can be calculated using the latent heat formula:
Q = mL
where:
Q is the heat transferred
m is the mass of the substance
L is the latent heat of the substance
Let's calculate the mass of ice needed:
-35 cal = m × 79.7 cal/g
m = (-35 cal) / (79.7 cal/g) ≈ -0.44 g
Since we cannot add a negative mass of ice, we conclude that we don't need to add any ice to lower the temperature of the tea to 12°C.