Answer to Question #89273 in General Chemistry for Shay

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.