How much ice initially at -4 °C is needed to cool down 250 g of water at 80 °C to 0°C?
Take the specific heat capacity of ice to be 2060 J kg-1°C-1
ambiguous...
more ice...none of which melts
OR
less ice...all of which melts
OR
some ice...which partially melts
but they want exactly just enough.
Heat in = heat to raise ice to 0 from -4 (specific heat of ice * ice mass * 4) + heat to melt ice (heat of fusion of water * ice mass)
=
Heat out = specific heat water * 80 * mass of water
To find out how much ice initially at -4 °C is needed to cool down 250 g of water at 80 °C to 0°C, we can use the formula:
Q = mcΔT
Where:
Q is the amount of heat transferred
m is the mass of the substance
c is the specific heat capacity
ΔT is the change in temperature
First, let's calculate the amount of heat transferred from the water to the ice. The water needs to be cooled from 80 °C to 0 °C, so ΔT for water is 80 °C - 0 °C = 80 °C.
mass of water (m water) = 250 g
specific heat capacity of water (c water) = 4,186 J kg^(-1) °C^(-1) (approximately)
Q water = m water * c water * ΔT water
Q water = 250 g * 4,186 J kg^(-1) °C^(-1) * 80 °C
Now, let's calculate the amount of heat transferred to the ice to raise its temperature from -4 °C to 0 °C. The ice needs to be heated by 4 °C, so ΔT for ice is 4 °C.
mass of ice (m ice) = ?
specific heat capacity of ice (c ice) = 2060 J kg^(-1) °C^(-1)
Q ice = m ice * c ice * ΔT ice
Q ice = m ice * 2060 J kg^(-1) °C^(-1) * 4 °C
Since the amount of heat transferred from the water is equivalent to the amount of heat transferred to the ice, we can set the two equations equal to each other:
Q water = Q ice
250 g * 4,186 J kg^(-1) °C^(-1) * 80 °C = m ice * 2060 J kg^(-1) °C^(-1) * 4 °C
Now we can solve for m ice, the mass of ice required.
m ice = (250 g * 4,186 J kg^(-1) °C^(-1) * 80 °C) / (2060 J kg^(-1) °C^(-1) * 4 °C)
Calculating this equation gives the mass of ice needed to cool down the water.