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?

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

Substitute in variables [Specific Heat Formula]:                                             q = (6.00 g)(4.18 J/g K)(39.0 °C)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

The atomic nucleus made up from protons (yellow) and neutrons (blue) and held together by nuclear forces. nuclear forces (also known as nuclear interactions or strong forces) are the forces that act between two or more nucleons. they bind protons and neutrons (“nucleons”) into atomic

Heat capacity = mcδtheat capacity = 2.676×10^3 j ..the heat capacity given is for 7.00 g of kcn,potassium cyanide which is equal to 0. mole of mole = mass  ÷ mr of kcn is 7  ÷ 65 =0.therefore: 0. of kcn liberates 2.676×10^3 j of heat1 mole   (2.676×10^3    ÷  0.107 )  × 1 =do the calculation to obtain 25009 j = 25kj