A 210W heater is placed in 2KG of water.whattemperature rise is produced if the heater is switched on for 200s?
210*200 Joules=2000*4.186j/gC*change in temp
change in temp= 21/4.186 C
check math.
To calculate the temperature rise produced by a heater, we need to use the formula:
ΔQ = m * c * ΔT
Where:
ΔQ - the amount of heat transferred (in Joules)
m - the mass of the water (in kg)
c - the specific heat capacity of water (approximately 4186 J/kg°C)
ΔT - the change in temperature (in °C)
First, let's calculate the amount of heat transferred:
ΔQ = P * t
Where:
P - the power of the heater (in Watts)
t - the time it is switched on (in seconds)
Given:
P = 210 W
t = 200 s
ΔQ = 210 W * 200 s
ΔQ = 42000 J
Now, we can calculate the temperature rise:
42000 J = 2 kg * 4186 J/kg°C * ΔT
Dividing both sides by (2 kg * 4186 J/kg°C):
ΔT = 42000 J / (2 kg * 4186 J/kg°C)
ΔT ≈ 5.02 °C
Therefore, the temperature rise produced by the heater when switched on for 200 seconds is approximately 5.02 °C.