General Chemistry | Heat Capacity (q=smΔT) [Example 1]

The video tutorial explains the concept of heat capacity using the formula Q=SMΔT. It begins with a problem involving heating 50 grams of water from 25°C to 85°C, calculating the heat required. The tutorial details the variables involved, such as specific heat, mass, and temperature change, and demonstrates the calculation process. It also covers converting volume to mass using density, particularly for water. The video concludes with a preview of more complex problems involving algebra, to be covered in the next video.

What is the initial temperature of water in the problem discussed?

• 25°C

• 45°C

• 15°C

• 35°C

Which unit is typically used for specific heat in the heat capacity formula?

• Calories per gram

• Joules per gram per Kelvin

• Joules per liter

• Watts per meter

What is the specific heat of water used in the calculation?

• 5.186 J/g°C

• 6.186 J/g°C

• 4.186 J/g°C

• 3.186 J/g°C

How is the heat calculated in the problem converted from joules to kilojoules?

• Multiply by 1000

• Divide by 1000

• Add 1000

• Subtract 1000

What is the final temperature of water in the problem discussed?

• 85°C

• 105°C

• 75°C

• 95°C

What is the density of water assumed in the problem when converting volume to mass?

• 0.5 g/mL

• 1.0 g/mL

• 1.5 g/mL

• 2.0 g/mL

If given a volume of water, what additional information is needed to use the heat capacity formula?

• Density

• Viscosity

• Temperature

• Pressure