Join thousands of students who trust us to help them ace their exams!
Multiple Choice
Calculate the amount of heat (in J) required to heat 2.00 grams of ice from −12.0℃ to 88.0℃. The heat of fusion of water is 6.02 kJ/mol, the specific heat capacity of ice is 2.09 J/g ▪℃ and the specific heat capacity of water is 4.184 J/g ▪℃.
A
50.2 J
B
418 J
C
668 J
D
736 J
E
1454 J
0 Comments
Verified step by step guidance
1
First, calculate the amount of heat required to raise the temperature of the ice from -12.0°C to 0°C using the formula: \( q = m \cdot c \cdot \Delta T \), where \( m \) is the mass, \( c \) is the specific heat capacity of ice, and \( \Delta T \) is the change in temperature.
Next, calculate the heat required to melt the ice at 0°C to water at 0°C using the formula: \( q = n \cdot \Delta H_f \), where \( n \) is the number of moles of ice and \( \Delta H_f \) is the heat of fusion. Convert the mass of ice to moles using the molar mass of water (18.02 g/mol).
Then, calculate the amount of heat required to raise the temperature of the water from 0°C to 88.0°C using the formula: \( q = m \cdot c \cdot \Delta T \), where \( c \) is the specific heat capacity of water.
Add all the calculated heat values from the previous steps to find the total heat required to heat the ice from -12.0°C to 88.0°C.
Ensure all units are consistent, converting kJ to J where necessary, and sum the heat values to find the total heat in Joules.