At an elevation where the boiling point of water is 93°C, 1.00 kg of water at 30°C absorbs 290.0 kJ from a mountain climber’s stove. Is this amount of thermal energy sufficient to heat the water to its boiling point? [cp of water = 4.18 J/(g · °C)] need more information to calculate can not be calculated even with more information no yes

Yes, it will be enough.

Explanation:

We can calculate the heat (Q) required to heat 1.00 kg of water from 30°C to 93°C using the following expression.

Q = cp × m × ΔT

where,

Q = cp × m × ΔT

Q = 4.18 J/g°C × 1.00 × 10³ g × (93°C-30°C)

Q = 263 kJ

Since 263 kJ are necessary, 290.0 kJ will be enough to heat the water.

c (diamond),

sio2 (quartz)

6.4 cm (max) - 25 cm (min)