The specific heat of water is 4.18 J/gºC. How many Joules of
heat are released to lower the temperature of 150 g of water
from 100 °C to 5°C?
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Respuesta :

Answer:

Q = 59565 [J]

Explanation:

In order to calculate the amount of thermal energy needed we must use the following equation.

[tex]Q=m*C_{p}*(T_{i}-T_{f})[/tex]

where:

m = mass = 150 [g]

Cp = 4.18 [J/g*°C]

Tfinal = 5 [°C]

Tinitial = 100 [°C]

Now replacing:

[tex]Q=150*4.18*(100-5)\\Q=59565[J][/tex]