How much energy is needed to raise the temperature of 1500g of ice from -21.0 degrees Celsius to 113 degrees Celsius
To calculate the energy needed to raise the temperature of a substance, you can use the specific heat formula:
Q = mcΔT
Where:
Q is the heat energy
m is the mass of the substance
c is the specific heat capacity of the substance
ΔT is the change in temperature
In this case, we're dealing with ice. The specific heat capacity of ice is 2.09 J/g°C. However, we need to consider that there are two phases involved: the temperature change from -21.0°C to 0°C when the ice melts, and then the temperature change from 0°C to 113°C when the water reaches its final temperature.
Let's break down the calculation into two parts:
Part 1: Heating the ice to its melting point (0°C)
First, we calculate the energy required to raise the temperature of the ice from -21.0°C to 0°C.
Q1 = mcΔT1
Where:
m = mass of ice = 1500 g
c = specific heat capacity of ice = 2.09 J/g°C
ΔT1 = change in temperature = 0°C - (-21.0°C) = 21.0°C
Q1 = (1500 g) * (2.09 J/g°C) * (21.0°C)
Q1 = 69495 J
So, the energy required to heat the ice from -21.0°C to 0°C is 69495 Joules.
Part 2: Heating the water from 0°C to 113°C
Next, we calculate the energy required to raise the temperature of the water from 0°C to 113°C.
Q2 = mcΔT2
Where:
m = mass of water = 1500 g
c = specific heat capacity of water = 4.18 J/g°C (assumed at room temperature)
ΔT2 = change in temperature = 113°C - 0°C = 113°C
Q2 = (1500 g) * (4.18 J/g°C) * (113°C)
Q2 = 894570 J
So, the energy required to heat the water from 0°C to 113°C is 894570 Joules.
Finally, we add up the energy from both parts to get the total energy required:
Total energy = Q1 + Q2
Total energy = 69495 J + 894570 J
Total energy = 964065 J
Therefore, the total energy needed to raise the temperature of 1500g of ice from -21.0 degrees Celsius to 113 degrees Celsius is 964065 Joules.