How many grams of water can be heated 71 °C by the absorption of 970 Joules?

Answer:
m = 3.3 g
Explanation:
Hello!
In this case, since the specific heat of water is defined as 4.184 J(g*°C), given the temperature change and the absorbed energy, we can compute the mass of water as shown below:
[tex]Q=mC\Delta T\\\\m=\frac{Q}{C\Delta T}[/tex]
Thus, we plug in the given data to obtain:
[tex]m=\frac{970J}{4.184\frac{J}{g\°C}*71\°C}\\\\m= 3.3g[/tex]
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