How much heat is required to warm 1.30 l of water from 24.0 ∘c to 100.0∘c? (assume a density of 1.0g/ml for the water.)?

1 answer:

5 0

When q = M* C *ΔT

we can get the mass using the given volume & density:

M = volume * density

= 1.3 L * 1g/ml * 1000mL1L

= 1300 g
and when the specific heat of water = 4.18J/g.°C

now, by substitution:

∴ q = 1300g * 4.18J/g.°C * (100-24)

= 412984 J or 412.984 KJ

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