temperature

logistic_guy

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How much heat (joules) is required to raise the temperature of \(\displaystyle 30.0 \ \text{kg}\) of water from \(\displaystyle 15^{\circ}\text{C}\) to \(\displaystyle 95^{\circ}\text{C}\)?
 
We use the heat transfer equation to solve this problem.

\(\displaystyle \Delta Q = mc\Delta T\)

where \(\displaystyle c\) is the specific heat capacity. For water \(\displaystyle c = 4186 \ \text{J/kg} \cdot \text{C}^{\circ}\)

Then,

\(\displaystyle \Delta Q = (30)(4186)(95 - 15) = \textcolor{blue}{1 \times 10^7 \ \text{J}}\)

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