Question

A quantity of water is heated from 25.0°C to 36.4°C by absorbing 325 J of heat energy. If the specific heat of water is 4.18 J / g°C, what mass is this quantity of water?

Answer 1

Answer:

6,8 g

Explanation:

c = 4.18 J/(g * °C) = 4180 J / (kg * °C)

$t_{1}$ = 25 °C

$t_{2}$ = 36,4 °C

Q = 325 J

The formula is: Q = c * m * ($t_{2} - t_{1}$)

m = $\frac{Q}{c * (t_{2} - t_{1} )}$

Calculating:

m = 325 / 4180 * (36,4 - 25) ≈ 0,0068 kg = 6,8 g