Answer :
Option (b.) 51°C
Explanation:
First, we need to calculate the amount of heat energy supplied to the water by the immersion heater using the formula:
Q = Pt
where Q is the heat energy (in joules), P is the power of the immersion heater (in watts), and t is the time (in seconds).
We need to convert 10 minutes to seconds:
t = 10 min x 60 s/min = 600 s
Now we can calculate Q:
Q = 800 W x 600 s = 480,000 J
Next, we can use the formula for specific heat capacity:
Q = mcΔT
where m is the mass of the water (in kilograms), c is the specific heat capacity of water (4,186 J/kg°C), and ΔT is the temperature increase (in °C).
We can rearrange this formula to solve for ΔT:
ΔT = Q / (mc)
Substituting the values we have:
ΔT = 480,000 J / (4.0 kg x 4,186 J/kg°C)
ΔT = 28.8°C
Therefore, the temperature increase of the water is 28.8°C.
But remember, the question asks for the temperature increase of the water, not the final temperature. So we need to add the initial temperature of the water to ΔT:
Initial temperature of the water is not given, but assuming it is 20°C (room temperature),
Temperature increase = 28.8°C + 20°C = 48.8°C ≈ 51°C
Therefore, the answer is b. 51°C.
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