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If there are 42.7 mL of water that change from 2.5°C to 36.8°C, calculate the number of joules of energy transferred in the form of heat

User CarlosV
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1 Answer

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Sure, to solve this question, we must first calculate the change in temperature, which we can get by subtracting the initial temperature from the final temperature.

So let's go step by step:

1. First, we have to find the change in temperature (`delta_temp`), which is done by subtracting initial temperature from the final temperature:

>> Final temperature = 36.8°C
>> Initial temperature = 2.5°C
So, `delta_temp` = Final temperature - Initial temperature
= 36.8°C - 2.5°C
= 34.3°C

Next, we are going to calculate the energy transferred in the process. To calculate the energy transferred, the formula used is q = m * c * deltaT, where

m = mass of the substance, which is the amount of water we have (in grams)
c = specific heat capacity of the substance (for water, it's approximately 4.186 joules/gram °C)
deltaT = change in temperature

2. Substituting the known values into the formula, we get:

>> Mass of water, m = 42.7 g
This is because 1 mL of water weighs approximately 1 gram, so for 42.7 mL of water, it's 42.7 grams.

>> Specific heat capacity, c = 4.186 joule/gram °C

Then we substitute all these into the formula:

q = m * c * deltaT # where q is the energy transferred
= 42.7g * 4.186 joule/gram °C * 34.3°C
= 6130.86 joules

So, the change in temperature is 34.3°C and the energy transferred is 6130.86 joules. This is the amount of energy that was transferred for the water to experience this temperature change.

User Rasika
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