Question

An unknown solid with a mass of 2.00 kilograms remains in the solid state while it absorbs 32.0 kilojoules of heat. Its temperature rises 4.00 degrees Celsius. What is the specific heat of the unknown solid?

Answer 1

Answer: The specific heat of the unknown solid is $4.00J/g^0C$

Explanation:

As we know that,  

$q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$    (1)

where,

q = heat absorbed  = 32.0 kJ = $32.0\times 10^3J$ J      (1kg=1000g)

$m$ = mass of unknown solid= 2.00 kg  = $2.00\times 10^3g$ (1kg=1000g)

$T_{final}$ = final temperature

$T_{initial}$ = initial temperature

$\Delta T$ =$4.00^0C$

$c$ = specific heat of unknown solid = ?

Now put all the given values in equation (1), we get

$32.0\times 10^3J=2.00\times 10^3g\times c\times (4.00^0C)]$

$c=4.00J/g^0C$

Therefore, the specific heat of the unknown solid is $4.00J/g^0C$

Answer 2

The scientific heat of the unknown solid will be "4.00 J/g°C".

Specific heat:

Given values are:

Heat absorbed, q = 32.0 kJ or, $32.0\times 10^3$ J

Mass, m = 2.00 kg or, $2.00\times 10^3$ g

Rise in temperature, ΔT = 4.00°C

We know the relation,

→ q = m×c×ΔT

or,

→    = m×c×($T_{final} - T_{initial}$)

By substituting the values,

$32.0\times 10^3=2.00\times 10^3\times c\times 4.00$

             $c = 4.00$ J/g°C    

Thus the above answer is appropriate.

Find out more information about specific heat here:

https://brainly.com/question/1768769