Question

How many joules of heat are absorbed to raise the temperature of 435 grams of water at 1 atm from 25 degrees celcius to its boiling point 100 degrees celcious?

Answer 1

In order to calculate the energy required, we must first know the specific heat capacity, or the amount of energy required to raise the temperature of a unit mass of substance by 1 degree Celsius, of water. This, at 1 atm pressure, is 4.18 joules per gram.
Next, we use the formula:
Q = mcΔT, where Q is the energy, m is the mass, c is the specific heat capacity and ΔT is the change in temperature.
Q = 435 * 4.18 * (100 - 25)
Q = 136.4 kJ

Answer 2

$\boxed{136568.25{\text{ J}}}$ of heat is absorbed to raise the temperature of 435 g of water from $25^\circ\text{C}$  to $100{\text{ }}^\circ {\text{C}}$.

Further explanation:

Properties of any substance are categorized as follows:

1. Intensive properties:

The properties that depend only on the nature of the substance and not on the size of the system are known as intensive properties. Their values remain unchanged even if a system gets divided into several subsystems. Temperature, refractive index, concentration, pressure, and density are some of the examples of intensive properties.

2. Extensive properties:

The properties that depend on the amount of the substance are called extensive. Their values get added up if a system is divided into several subsystems. Mass, enthalpy, volume, energy, size, weight, and length are some of the examples of extensive properties.

Specific heat

The amount of heat required to increase the temperature of a substance per unit mass. Its SI unit is Joule (J).

The expression for the amount of heat of water is as follows:

${\text{Q}} = \text{mc}\;\Delta T}}$        …… (1)

                                                                                         

Here,

Q is the amount of heat absorbed.

m is the mass of water.

c is the specific heat of the water.

$\Delta\text{T}$ is the change in temperature of the water.

The formula to calculate the change in temperature of the water is as follows:

$\Delta\text{T} = {\text{Final temperature of water}} - {\text{Initial temperature of water}}$        …… (2)

                           

The value of the initial temperature of water is $25{\text{ }}^\circ {\text{C}}$.

The value of the final temperature of water is ${\text{100 }}^\circ {\text{C}}$.

Substitute these values in equation (2).

$\begin{aligned}\Delta \text{T} &= {\text{100 }}^\circ {\text{C}} - 25{\text{ }}^\circ {\text{C}} \\&= {\text{75 }}^\circ {\text{C}}\\\end{aligned}$

 

The value of $\Delta\text{T}$ is $75^\circ\text{C}$.

The value of m is 435 g.

The value of c is $4.186\;{\text{J/g}}\;^\circ\text{C}$.

Substitute these values in equation (1).

$\begin{aligned}{\text{Q}} &= \left( {{\text{435 g}}} \right)\left( {4.186\;{\text{J/g}}\;^\circ\text{C} \right)\left( {75{\text{ }}^\circ {\text{C}}} \right)\\ &= 136568.25{\text{ J}}\\\end{aligned}$

Learn more:

1. Calculate the enthalpy change using Hess’s Law: brainly.com/question/11293201
2. Find the enthalpy of decomposition of 1 mole of MgO: brainly.com/question/2416245

Answer details:

Grade: Senior School

Chapter: Thermodynamics

Subject: Chemistry

Keywords: intensive, extensive, specific heat, m, c, Q, 435 g, 136568.25 J, water, mass of water, initial temperature, final temperature, heat absorbed.