1) what mass of ice at -14'c will be needed to cool 200cm3 of an orange drink (essentially water) from 25'c to 10'c!

(specific latent heat of fusion of ice =336000)
(specific heat capacity of ice = 2100)
(specific heat capacity of water = 4200)

0.492

To find the mass of ice needed to cool the orange drink, we need to consider the heat gained by the ice and the heat lost by the orange drink.

First, let's calculate the heat lost by the orange drink using the formula:

Q = mcΔT

Where:
Q - heat lost by the orange drink
m - mass of the orange drink
c - specific heat capacity of water
ΔT - change in temperature of the orange drink

Given that the mass of the orange drink is 200 cm³ and the specific heat capacity of water is 4200 J/kg°C, we need to convert the volume to mass. The density of water is 1 g/cm³, so:

200 cm³ × 1 g/cm³ = 200 g

Therefore, the mass of the orange drink is 200 grams.

Now, let's calculate the heat lost by the orange drink:

Q = (200 g) × (4200 J/kg°C) × (25°C - 10°C)
Q = 200 g × 4200 J/kg°C × 15°C
Q = 1,260,000 J

The heat lost by the orange drink is 1,260,000 J.

Next, let's calculate the heat gained by the ice when it melts using the formula:

Q = mL

Where:
Q - heat gained by the ice
m - mass of the ice
L - specific latent heat of fusion of ice

Given that the specific latent heat of fusion of ice is 336,000 J/kg, we need to find the mass of the ice.

We can use the formula:

Q = mL
m = Q / L

m = 1,260,000 J / 336,000 J/kg
m ≈ 3.75 kg

Therefore, approximately 3.75 kg of ice is needed to cool the orange drink to the desired temperature.

Note: Remember that the temperature of the ice must be -14°C in order to cool the drink to 10°C.