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

Answer 2

The mass of the quantity of water heated from 25.0°C to 36.4°C by absorbing 325 J of heat energy is 6.8grams.

HOW TO CALCULATE MASS:

The mass of a substance can be calculated by using the following formula:

Q = m × c × ∆T

Where;

• Q = quantity of heat absorbed (J)
• c = specific heat capacity (4.18 J/g°C)
• m = mass of substance
• ∆T = change in temperature (°C)

According to this question;

m = ?

∆T = 36.4°C - 25°C = 11.4°C

Q = 325J

325 = m × 4.18 × 11.4

325 = 47.65m

m = 325 ÷ 47.65

m = 6.8g

Therefore, the mass of the quantity of water heated from 25.0°C to 36.4°C by absorbing 325 J of heat energy is 6.8grams.

Learn more about specific heat capacity at: https://brainly.com/question/2530523?referrer=searchResults