Question

450. grams of water are heated from 20.0 °C to 37.0 °C. How many kcal were absorbed by the water? Note that this is the amount of heat absorbed when a glass of cold water is consumed.​

Answer 1

Answer:

Q = 32,007.6 J

Explanation:

Hello there!

In this case, since the energy involved during a heating/cooling process is:

$Q=mC(T_f-T_i)$

Thus, given the mass, specific heat of water, initial temperature and final one, we plug in obtain:

$Q=450g*4.184\frac{J}{g\°C} (37.0\°C-20.0\°C)\\\\Q=32,007.6J$

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