Question

A sample of water (88 grams) had a temperature change of 6.0 ℃. What was the heat change of this sample? (q = m c ΔT, c = 4.18 J/ g ℃)

Answer 1

Answer:

$Q=2209J$

Explanation:

Hello!

In this case, since the heat involved during a heating process is computed in terms of mass, specific heat and temperature change as shown below:

$Q=mC\Delta T$

Thus, since the heated mass of water was 88 g, the specific heat of water is 4.184 J/g°C and the temperature change is 6.0 °C, we can compute the heat as shown below:

$Q=88g*4.184\frac{J}{g\°C}*6.0\°C \\\\Q=2209J$

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