Question

The specific heat capacity of liquid water is 4.18 J/g-K. How many joules of heat are needed to raise the temperature of 6.00 g of water from 36.0°C to 75.0 °C?

Answer 1

Answer:

978 J

General Formulas and Concepts:

Thermodynamics

Specific Heat Formula: q = mcΔT

• q is heat (in J)
• m is mass (in g)
• c is specific heat (in J/g °C)
• ΔT is change in temperature (in °C)

Explanation:

Step 1: Define

[Given] c = 4.18 J/g K

[Given] m = 6.00 g

[Given] ΔT = 75.0 °C - 36.0 °C = 39.0 °C

[Solve] q

Step 2: Find q

1. Substitute in variables [Specific Heat Formula]:                                             q = (6.00 g)(4.18 J/g K)(39.0 °C)
2. Multiply [Cancel out units]:                                                                              q = 978.12 J

Step 3: Check

Follow sig fig rules and round. We are given 3 sig figs.

978.12 J ≈ 978 J

Answer 2

Answer:

I don't remember this topic very well.

Explanation:

if we want to increase 1C° of 1 gram water. we must use 4.18 joules energy that 75-36= 39 C° for 6 gram H2O we should use 6*39*(4.18) joules