Question

How much heat is required to warm 1.20 l of water from 26.0 ∘c to 100.0 ∘c? (assume a density of 1.0g/ml for the water.)?

Answer 1

Answer: The amount of heat required is 371.72 kJ

Step-by-step explanation:

To calculate the mass of water, we use the equation:

`\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}`

Density of water = 1 g/mL

Volume of water = 1.20 L = 1200 mL (Conversion factor: 1 L = 1000 mL)

Putting values in above equation, we get:

`1g/mL=\frac{\text{Mass of water}}{1200mL}\\\\\text{Mass of water}=(1g/mL* 1200mL)=1200g`

To calculate the amount of heat absorbed or released, we use the equation:

`Q= m* c* \Delta T`

Q = heat absorbed = ?

m = mass of water = 1200 g

c = specific heat capacity of water = 4.186 J/g °C

`\Delta T={\text{Change in temperature}}=(100-26)^oC=74^oC`

Putting values in above equation, we get:

`Q=1200g* 4.186J/g^oC* 74^oC=371716.8J=371.72kJ`

Conversion factor used: 1 kJ = 1000 J

Hence, the amount of heat required is 371.72 kJ

Answer 2

when we have the volume = 1.2 L

and density = 1000 g/L

∴ mass of water(m) = volume * density

= 1.2 * 1000

= 1200 g

according to this formula, we can get the amount of Heat q:


q = m *C* ΔT

= 1200 g * 4.18 * (100 - 26)

= 371184 J