$1.2 xx 10^5$ $J$ of thermal energy are added to a sample of water and its temperature changes from $229$ $K$ to $243$ $K$ . What is the mass of the water?

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$1.2 xx 10^5$ $J$ of thermal energy are added to a sample of water and its temperature changes from $229$ $K$ to $243$ $K$ . What is the mass of the water?

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Answer 1 · 2016-03-23T23:45:09

$Delta H = mCDeltaT$

Rearranging:

$m = (Delta H)/(CDeltaT)=(1.2xx10^5)/(4181xx(243-229))=2.1$ $kg$

Explanation:

We use the equation $Delta H = mCDeltaT$ where $m$ is the mass $(kg)$ , $T$ is the temperature $(K)$ , $H$ is the thermal energy $(J)$ and $C$ is the specific heat $(Jkg^-1K^-1)$ . $Delta$ just means 'the change in'.

We know that for water, $C=4181$ $Jkg^-1K^-1$ . (Sometimes quoted as $4.2$ $Jg^-1K^-1$ , but it's better to work in SI units.)

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$1.2 xx 10^5$ $J$ of thermal energy are added to a sample of water and its temperature changes from $229$ $K$ to $243$ $K$ . What is the mass of the water?

1 Answer

Explanation:

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1.2 xx 10^51.2 xx 10^5 JJ of thermal energy are added to a sample of water and its temperature changes from 229229 KK to 243243 KK. What is the mass of the water?

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$1.2 xx 10^5$ $J$ of thermal energy are added to a sample of water and its temperature changes from $229$ $K$ to $243$ $K$ . What is the mass of the water?