How much heat (in kj) is needed to convert 856 g of ice at −10.0°c to steam at 126.0°c? (the specific heats of ice, water, and steam are 2.03 j/g · °c, 4.184 j/g · °c, and 1.99 j/g · °c, respectively. the heat of fusion of water is 6.01 kj/mol, the heat of vaporization is 40.79 kj/mol.)

The heat required for the given process is 2659.3 kJ

Explanation:

The processes involved in the given problem are:  

Pressure is taken as constant.

To calculate the amount of heat absorbed at different temperature, we use the equation:

       .......(1)

where,

q = amount of heat absorbed = ?

= specific heat capacity of medium

m = mass of water/ice

= final temperature

= initial temperature

To calculate the amount of heat released at same temperature, we use the equation:

      ......(2)

where,

q = amount of heat absorbed = ?

m = mass of water/ice

= latent heat of fusion or vaporization

Calculating the heat absorbed for each process:

For process 1:

We are given:

Putting values in equation 1, we get:

For process 2:

Converting the latent heat of fusion in J/g, we use the conversion factor:

Molar mass of water = 18 g/mol

1 kJ = 1000 J

So,

We are given:

Putting values in equation 2, we get:

For process 3:

We are given:

Putting values in equation 1, we get:

For process 4:

Converting the latent heat of vaporization in J/g, we use the conversion factor:

Molar mass of water = 18 g/mol

1 kJ = 1000 J

So,

We are given:

Putting values in equation 2, we get:

For process 5:

We are given:

Putting values in equation 1, we get:

Total heat absorbed =

Total heat absorbed =

Hence, the heat required for the given process is 2659.3 kJ

4- c, because all matter stays the same no matter what. it is still there. 5- b, because matter is known as mass. 6-c because, flammability is a chemical property. malleability is changing the shape of an object.

Sorry i dont understand the question can you rephase it?