How many joules of heat are required to heat 116.4 grams of room temperature water (22°c) to the boiling point?

The amount of heat required to raise 116.4 g of water from 22°C to boiling point is 3.8 ×10^4 J (or around 38,000 J).  The boiling point of water is 100°C and the specific heat capacity of water is a constant, which is 4.18 J/g-°C.

Further Explanation:

The amount of heat absorbed related to a temperature change is determined using the equation:

where:

q = amount of heat absorbed (in Joules)

m = mass of the substance (in grams)

C = specific heat capacity, (in J/g-°C)

T(f) = final temperature

T(i) = initial temperature

The specific heat capacity, C, is the amount of heat required to raise the temperature by 1 °C of one gram of a substance. Tables listing the specific heat capacities of substances may be found in most chemistry textbooks and handbooks. However, it is useful to know that for water, the specific heat capacity is 4.18 J/g-°C.

To solve for q, identify the values  for m, C, T(f) and T(i), then insert into the equation.

m = 116.4 g (4 significant figures)

C = 4.18 J/g-°C

T(f) = boiling point of water = 100°C

T(i) = 22°C (2 significant figures)

Since the least number of significant figures in the given is 2, then the final answer must have 2 significant figures as well. Hence,

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Keywords: heat capacity, Joules, calorimetry

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explanation: