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Calculate the amount of heat needed to warm 16.1 kg of water from a room temperature of 20.0°C to 60.0°C. Show your work

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Answer:

The specific heat capacity of water is 4.18 J/g°C. We have 16.1 kg of water, which is equivalent to 16,100 g.

The change in temperature is:

ΔT = (final temperature) - (initial temperature) = 60.0°C - 20.0°C = 40.0°C

The amount of heat needed is given by:

q = m * c * ΔT

where q is the amount of heat needed, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature.

Plugging in the values we have:

q = 16,100 g * 4.18 J/g°C * 40.0°C

q = 2,697,880 J or 2.70 MJ (rounded to two significant figures)

Therefore, the amount of heat needed to warm 16.1 kg of water from 20.0°C to 60.0°C is approximately 2.70 MJ.

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