how many kj will be absorbed in changing 93kj to it at 20degree celsius to water in 40degree celsius show working
To calculate the amount of heat absorbed when changing 93 kJ of ice at 20 degrees Celsius to water at 40 degrees Celsius, you need to consider two steps: first, raising the temperature of the ice to its melting point, and second, melting the ice into water and heating it further to the final temperature.
Step 1: Raising the temperature of ice to its melting point.
The specific heat capacity of ice is 2.09 J/g°C.
First, calculate the amount of heat required to raise the temperature of the ice from 20°C to 0°C:
q1 = mass of ice * specific heat capacity of ice * temperature change
Assuming the mass of the ice is 1 gram:
q1 = 1 g * 2.09 J/g°C * (0°C - 20°C)
q1 = 1 g * 2.09 J/g°C * (-20°C)
q1 = -41.8 J
Step 2: Melting the ice into water and heating it further.
The heat of fusion (enthalpy of fusion) for ice is 334 J/g.
Calculate the heat required to melt the ice:
q2 = mass of ice * heat of fusion
q2 = 1 g * 334 J/g
q2 = 334 J
Next, calculate the heat required to raise the temperature of the water from 0°C to 40°C:
The specific heat capacity of water is 4.18 J/g°C.
q3 = mass of water * specific heat capacity of water * temperature change
Assuming the mass of water is also 1 gram:
q3 = 1 g * 4.18 J/g°C * (40°C - 0°C)
q3 = 1 g * 4.18 J/g°C * 40°C
q3 = 167.2 J
Finally, add up the three quantities of heat to get the total heat absorbed:
Total heat absorbed = q1 + q2 + q3
Total heat absorbed = -41.8 J + 334 J + 167.2 J
Total heat absorbed = 459.4 J
Therefore, the amount of heat absorbed in changing 93 kJ of ice at 20 degrees Celsius to water at 40 degrees Celsius is 459.4 J.