Question

How much heat must be absorbed by 375 grams of water to raise its temperature by 25 degrees celsius? The specific heat of water is 4.186 J/ g degrees Celsius

Answer 1

$Q=mc \Delta T$

$Q$ - heat content

$m$ - mass

$c$ - specific heat

$\Delta T$ - temperature change

$Q = 375 \cdot 4.186 \cdot 25 \approx 3.92 \cdot 10^4 \; J = 39.2 \; kJ$

Answer: 39.2 kJ.