Question

A beaker contains 0.500 kg of water at a temperature of 3.0 °C. The beaker is heated, and the
internal energy of the water increases by 21.0 kJ.
The specific heat capacity of water is 4200 J / (kg °C).
What is the temperature of the water after it has been heated?

Answer 1

Answer:

13°C

Explanation:

We can apply the formula for Heat energy gained by a body. It is given as:

$H = mc(T_2 - T_1)$

where m = mass of the body

c = Specific heat capacity of the body

$T_2$ = final temperature of the body

$T_1$ = initial temperature of the body

For the water in the beaker, we are given that:

m = 0.5 kg

$T_1$  = 3.0 °C

H = 21 kJ = 21000 J

c = 4200 J/kg°C

Hence, the final temperature $T_2$ of the water is:

21000 = 0.5 * 4200 * ($T_2$ - 3)

21000 = 2100 * ($T_2$  - 3)

($T_2$  - 3) = $\frac{21000}{2100}$

$T_2$  - 3 = 10

$T_2$  = 10 + 3 = 13°C

The final temperature of the water is 13°C.

Answer 2

Answer:

The temperature of the water after it has been heated is 13°C

Explanation:

Heat capacity H is expressed as shown :

H = mc∆t

H is the internal energy of the water

m is the mass of the water

c is the specific heat capacity of the water

∆t is the change temperature

Given H = 21kJ = 21,000Joules

m = 0.500kg

c = 4200 J / (kg °C).

∆t = t2-t1

t2 is the final temperature

t1 is the initial temperature = 3°C

Substituting the parameters;

21000 = 0.5(4200)(t2- 3)

21000 = 2100(t2-3)

21000 = 2100t2 - 6300

2100t2 = 21000+6300

2100t2 = 27300

t2= 27300/2100

t2 = 13°C