What quantity of heat (kJ) is required to convert 200g of ice at 0.00C into water at 50C?
Heat=mass*Heatfusion + mass*specificheatcapacity*50
To solve this problem, we need to calculate the heat required to change the temperature of the ice from 0.00°C to its melting point (0.00°C) and then the heat required to change the ice into liquid water, and finally, the heat required to change the temperature of the water from 0.00°C to 50.00°C.
First, we calculate the heat required to raise the temperature of the ice from 0.00°C to its melting point:
q1 = m * Cp * ΔT
where:
- m is the mass of the ice (200g)
- Cp is the specific heat capacity of ice, which is 2.09 J/g°C
- ΔT is the change in temperature (0.00°C - 0.00°C = 0°C)
So, q1 = 200g * 2.09 J/g°C * 0°C = 0 J
Next, we calculate the heat required to convert the ice into liquid water:
q2 = m * ΔHfus
where:
- ΔHfus is the heat of fusion or the enthalpy of fusion which is the heat required to change a substance from a solid to a liquid state. For water, this value is 334 J/g.
So, q2 = 200g * 334 J/g = 66,800 J
Lastly, we calculate the heat required to raise the temperature of the liquid water from 0.00°C to 50.00°C:
q3 = m * Cp * ΔT
where:
- m is the mass of the liquid water, which is the same as the mass of the ice (200g)
- Cp is the specific heat capacity of water, which is 4.18 J/g°C
- ΔT is the change in temperature (50.00°C - 0.00°C = 50.00°C)
So, q3 = 200g * 4.18 J/g°C * 50.00°C = 41,800 J
Now, to find the total heat required, we add up the values obtained from each step:
q_total = q1 + q2 + q3 = 0 + 66,800 J + 41,800 J = 108,600 J
To convert this value to kilojoules (kJ), we divide by 1000:
q_total_kJ = 108,600 J / 1000 = 108.6 kJ
Therefore, the quantity of heat required to convert 200g of ice at 0.00°C into water at 50.00°C is 108.6 kJ.