Answer:
42.01 °C
Explanation:
<u>Step 1: explain the problem</u>
We have to find the initial temperature, when a certain amount of heat raises this sample of water to its boiling point ( 100 °C)
⇒this amount of heat = 5.51 x 10Δ^5 J
We will use the formule : Q = mcΔ
T
with Q = heat transfer ( J)
with m = mass of the substance (g)
with c = specific heat ( J/g °C)
with Δ
T = temperature change ( in °C or K)
Water has a specific heat of 4.186 J/g °C
<u>Step 2 : Calculate the initial temperature </u>
We have to rearrange the formule first:
Δ
T = Q / mc
In this case we have :
Δ
T = 5.51 * 10^5 J / 2270 g * 4.186 J/g °C = 57.99
⇒ The final temperature of the water is the boiling point (100 °C) and the change of temperature is 57.99. This means that the boiling point is 57.99 °C higher than the initial temperature.
This means : Δ
T = Tboiling point - Tinitial
Δ
T = 57.99 °C = 100 °C - Tinitial
Tinitial = 100 °C - 57.99 °C = 42.01 °C
The initial temperature of the water is 42.01 °C
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