A 210W heater is placed in 2KG of water. What temperature rise is produced if the heater is switched on for 200 seconds?
So without converting to calories the answer would be 21 degree C, right?
answer
To determine the temperature rise produced by the heater, we need to use the formula:
Q = mcΔT
Where:
Q represents the amount of heat energy transferred
m represents the mass of the water
c represents the specific heat capacity of water
ΔT represents the change in temperature
First, let's calculate the amount of heat energy transferred (Q) using the formula:
Q = P × t
Where:
P represents the power of the heater
t represents the time the heater is switched on
Substituting the given values:
P = 210W
t = 200 seconds
Q = 210W × 200 seconds
Q = 42,000 joules (J)
Next, we need to determine the specific heat capacity of water (c), which is approximately 4.18 J/g°C.
Now, we can rearrange the formula to solve for the change in temperature (ΔT):
ΔT = Q / (m × c)
Substituting the values:
Q = 42,000 J
m = 2,000 g (since 1 kg = 1000 g)
c = 4.18 J/g°C
ΔT = 42,000 J / (2,000 g × 4.18 J/g°C)
ΔT ≈ 5.03 °C
Therefore, the temperature will rise by approximately 5.03 degrees Celsius.
delta T = (heat energy added)/(M*C)
= (210 J/s*200s)/(M*C)
= 4.2*10^4 J/(M*C)
= 10,050 calories/(2 kg*1000 cal/kgC)
= 5.0 degrees
I converted to calories so I could use the easily remembered specific heat of water (C) in calorie/kg*C .