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What is the temperature increase of 4.0 kg of water when heated by an 800-W immersion heater for 10 minutes? (c_w = 4,186 J/kg⋅°C)

A. 56°C
B. 51°C
C. 29°C
D. 14°C

Answer :

Answer:

option C

Explanation:

given,

mass of water = 4 Kg

Water is heated to = 800 W

time of immersion = 10 min

= 10 x 60 = 600 s

using equation of specific heat

Q = m S ΔT

S is the specific heat capacity of water which is equal to 4182 J/kg°C.

and another formula of heat

Q = Pt

now,

P t = m S ΔT

800 x 600 = 4 x 4182 x ΔT

ΔT = 29° C

temperature increased is equal to ΔT = 29° C

Hence, the correct answer is option C

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Rewritten by : Barada

The correct option is c). The temperature increase is approximately 29°C.

To determine the temperature increase of 4.0 kg of water when heated by an 800-W immersion heater for 10 minutes, we can use the formula:

[tex]\[ Q = mc\Delta T \][/tex]

where:

- Q is the heat energy supplied (in joules)

- m is the mass of the water (in kilograms)

- c is the specific heat capacity of water (in J/kg⋅°C)

- [tex]\( \Delta T \)[/tex] is the temperature increase (in °C)

First, calculate the total heat energy supplied by the immersion heater:

[tex]\[ Q = \text{Power} \times \text{time} \][/tex]

The power of the heater is 800 W, and the time is 10 minutes (which is 600 seconds):

[tex]\[ Q = 800 \, \text{W} \times 600 \, \text{s} \][/tex]

[tex]\[ Q = 480,000 \, \text{J} \][/tex]

Now, use the specific heat capacity of water [tex](\( c = 4186 \, \text{J/kg} \)[/tex] °C)

[tex]\[ 480,000 \, \text{J} = 4.0 \, \text{kg} \times 4186 \, \text{J/kg} \times \Delta T \][/tex]

Solve for :

[tex]\[ \Delta T = \frac{480,000 \, \text{J}}{4.0 \, \text{kg} \times 4186 \, \text{J/kg}} \][/tex]

[tex]\[ \Delta T = \frac{480,000}{16,744} \][/tex]

[tex]\[ \Delta T = 28.67[/tex]°C

Rounding to the nearest whole number which is 29°C