A 5-kg block of ice is at a temperature of -27 °C. How much heat must be added to the
ice to produce 5 kg of liquid water with a final temperature of 35 oC?
Q=Q₁+Q₂+Q₃=
=mc(ice)(27-0)+mλ(ice) +mc(water) (35-0)=
=5(2060•27 + 335000 + 4183•35) = …
By the way, ice and water have different specific heat capacities. That is what was tripping me up.
To calculate the amount of heat required to convert the ice into liquid water, we need to consider two steps:
Step 1: Calculate the heat required to raise the temperature of the ice from -27 °C to 0 °C.
Step 2: Calculate the amount of heat required to convert the ice at 0 °C into liquid water at 35 °C.
Let's break down these steps.
Step 1: Calculate the heat required to raise the temperature of ice to 0 °C.
The specific heat capacity of ice, denoted as c, is 2.09 J/g°C. Given that the mass (m) of the ice is 5 kg (5000 g) and the initial temperature (T1) is -27 °C, we can calculate the heat energy (Q1) required using the formula:
Q1 = m * c * (T1 - 0)
Q1 = 5000 g * 2.09 J/g°C * (-27 - 0)
Q1 = -225,450 J
Note: We ignore the negative sign as it indicates the heat is being added to the system.
Step 2: Calculate the heat required to convert the ice at 0 °C into liquid water at 35 °C.
The heat of fusion (Hf) for water is 334 J/g. Since we have 5 kg (5000 g) of ice, the amount of heat energy required to convert it into freshwater at 0 °C is:
Q2 = m * Hf
Q2 = 5000 g * 334 J/g
Q2 = 1,670,000 J
Next, we need to calculate the heat energy required to raise the temperature of the liquid freshwater from 0 °C to 35 °C. The specific heat capacity of water (c) is 4.18 J/g°C. Therefore, the heat energy required is:
Q3 = m * c * (T2 - 0)
Q3 = 5000 g * 4.18 J/g°C * (35 - 0)
Q3 = 732,500 J
Finally, to find the total amount of heat energy required to convert the ice into liquid water at 35 °C, we add up Q1, Q2, and Q3:
Total heat required = Q1 + Q2 + Q3
Total heat required = -225450 J + 1670000 J + 732500 J
Total heat required = 1,369,050 J
Therefore, approximately 1,369,050 J (or 1,369.05 kJ) of heat must be added to the ice to produce 5 kg of liquid water with a final temperature of 35 °C.
To determine the amount of heat required to melt the ice and raise the temperature of the resulting water, we need to use the specific heat capacity and the heat of fusion of water. Here's the step-by-step process:
1. Calculate the heat required to raise the temperature of the ice to its melting point, 0 °C.
- The specific heat capacity of ice is 2.09 J/g·°C.
- The initial temperature is -27 °C.
- The mass of the ice is 5 kg (which is equivalent to 5000 g).
The formula for heat is Q = m·c·ΔT, where Q is the heat, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
Q1 = 5000 g × 2.09 J/g·°C × (0 °C - (-27 °C))
Q1 = 5000 g × 2.09 J/g·°C × 27 °C
Q1 = 283,950 J
2. Calculate the heat required to melt the ice at 0 °C.
- The heat of fusion of water is 334 J/g.
- The mass of the ice is still 5 kg (5000 g).
Q2 = 5000 g × 334 J/g
Q2 = 1,670,000 J
3. Calculate the heat required to raise the temperature of the resulting water from 0 °C to 35 °C.
- The specific heat capacity of liquid water is 4.18 J/g·°C.
- The mass of the water is 5 kg (5000 g).
Q3 = 5000 g × 4.18 J/g·°C × (35 °C - 0 °C)
Q3 = 5000 g × 4.18 J/g·°C × 35 °C
Q3 = 732,500 J
4. Add up the three heats to find the total heat required.
Q_total = Q1 + Q2 + Q3
Q_total = 283,950 J + 1,670,000 J + 732,500 J
Q_total = 2,686,450 J
Therefore, approximately 2,686,450 joules of heat must be added to the ice to produce 5 kg of liquid water with a final temperature of 35 °C.