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
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
