How much heat is required to warm 1.50 kg of sand from 20.0 C to 100.0 C?

q = mass sand x specific heat sand x 80

9600

To calculate the amount of heat required to warm the sand, we can use the equation:

Q = mcΔT

Where:
Q = Amount of heat (in joules)
m = Mass of the sand (in kg)
c = Specific heat capacity of the sand (in J/kg·°C)
ΔT = Change in temperature (in °C)

Given:
m = 1.50 kg
c = 830 J/kg·°C (approximately, for typical sand)
ΔT = 100.0°C - 20.0°C = 80.0°C

Substituting these values into the equation, we get:

Q = (1.50 kg)(830 J/kg·°C)(80.0°C)
Q = 99,600 J

Therefore, the amount of heat required to warm 1.50 kg of sand from 20.0°C to 100.0°C is 99,600 joules.

To calculate the amount of heat required to warm a substance, you can use the following equation:

Q = mcΔT

Where:
Q = Amount of heat transferred
m = Mass of the substance
c = Specific heat capacity of the substance
ΔT = Change in temperature

First, you need to find the specific heat capacity of sand. The specific heat capacity represents the amount of heat needed to raise the temperature of 1 kilogram of a substance by 1 degree Celsius. The specific heat capacity of sand is typically around 0.84 J/g°C.

Next, convert the mass of sand from kilograms to grams:

1.50 kg x 1000 g/kg = 1500 g

Now we can calculate the amount of heat:

Q = (mass) x (specific heat capacity) x (change in temperature)
Q = 1500 g x 0.84 J/g°C x (100.0°C - 20.0°C)

Q = 1500 g x 0.84 J/g°C x 80.0°C

Q ≈ 100,800 J

Therefore, the amount of heat required to warm 1.50 kg of sand from 20.0°C to 100.0°C is approximately 100,800 Joules.