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Andru [333]
2 years ago
6

A pan containing 20.0 grams of water was allowed to cool from a temperature of 95.0 °C. If the amount of heat released is 1,200

joules, what is the approximate final temperature of the water?
75 °C
78 °C
81 °C
87 °C
Chemistry
1 answer:
34kurt2 years ago
4 0

Answer:

81 °C

Explanation:

This is a calorimetry question so a few things you will need for this. The calorimetry equation q=mcΔT & the specific heat of water (4.2J/g•°C). Other definitions are:

q = heat added/released by a sample

m = mass of sample

c=specific heat of sample

ΔT = change in temperature

from here we can rearrange the equation to state:

q/(mc) = ΔT

1200J/((20.0g)(4.2J/g•°C)) = ΔT

14°C = ΔT

If the starting temperature was 95.0°C and we know that the temperature was cooled by 14°C then the final temperature of the water would be 81.

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<u>Given:</u>

Beaker 1:

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<u>To determine:</u>

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Final\ mass\ of \ water = Beaker\ 1 + Beaker\ 2 = 44.3 + 115.2 = 159.5 g

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Explanation:

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