1 litre of coffee at 80oC need to be cooled to 40oC. Assuming that all the ice melts and no heat is lost to the environment. Take the specific heat capacity of coffee to be 4200J/kgK, density of coffee to be 1000kg/m3 and latent heat of fusion of water to be 3.34 x 105 J/kg. You are required to:
a) Determine the heat lost by the coffee when ice is added and coffee cools to 40oC?
Solution:
To determine the heat lost by the coffee when ice is added and it cools to 40°C, we need to consider two processes:
1. Cooling the coffee from 80°C to 0°C.
2. Melting the ice and further cooling the coffee from 0°C to 40°C.
Let's calculate each process step by step:
1. Cooling the coffee from 80°C to 0°C:
The heat lost by the coffee can be calculated using the formula:
Q = mcΔT
Where Q is the heat lost, m is the mass of the coffee, c is the specific heat capacity of coffee, and ΔT is the change in temperature.
Given:
- Volume of coffee: 1 L
- Density of coffee: 1000 kg/m³ (since 1 L of coffee weighs 1000 kg)
- Specific heat capacity of coffee: 4200 J/kgK
- Initial temperature of coffee: 80°C
- Final temperature: 0°C
First, we need to calculate the mass of the coffee:
Mass = Volume × Density = 1 L × 1000 kg/m³ = 1000 kg
Next, we find the change in temperature:
ΔT = Final temperature - Initial temperature = 0°C - 80°C = -80°C
Now we can calculate the heat lost:
Q1 = mcΔT = 1000 kg × 4200 J/kgK × (-80°C) = -336,000,000 J
Note: The negative sign indicates heat lost by the coffee.
2. Melting the ice and further cooling the coffee from 0°C to 40°C:
To melt the ice, we need to calculate the heat required using the formula:
Q = ml
Where Q is the heat required, m is the mass of ice, and l is the latent heat of fusion of water.
Given:
- Latent heat of fusion of water: 3.34 × 10^5 J/kg (given)
The mass of ice can be calculated using the volume of the added ice and the density of ice, which is assumed to be the same as water:
Mass = Volume × Density = 1 L × 1000 kg/m³ = 1000 kg
Now we can calculate the heat required to melt the ice:
Q2 = ml = 1000 kg × 3.34 × 10^5 J/kg = 334,000,000 J
Next, we need to calculate the heat lost by the coffee when it cools from 0°C to 40°C:
Q3 = mcΔT = 1000 kg × 4200 J/kgK × (40°C - 0°C) = 168,000,000 J
Finally, we can calculate the total heat lost by the coffee:
Total heat lost = Q1 + Q2 + Q3 = -336,000,000 J + 334,000,000 J + 168,000,000 J = 166,000,000 J
Therefore, the heat lost by the coffee when ice is added and it cools to 40°C is 166,000,000 J.