a jug of heat capacity 250j/kg contains water at 28 degree celsius. an electric heater of resistance 35 holmes connected to 220 volt source is used to raise the temperature of water until it boils at 100 degree celcius in 4 minutes.find the mass of water in the jug before heating
Why did the water go to therapy? Because it had too many degrees!
To find the mass of water in the jug before heating, we need to use the equation:
Q = mcΔT
Where:
Q = heat energy transferred (Joules)
m = mass of the water (kg)
c = specific heat capacity of water (J/kg°C)
ΔT = change in temperature (°C)
In this problem, we know the following values:
Q = 250 J/kg
c = specific heat capacity of water = 4186 J/kg°C (approximate value)
ΔT = 100°C - 28°C = 72°C
Now we can rearrange the equation to solve for the mass:
m = Q / (cΔT)
m = 250 J/kg / (4186 J/kg°C * 72°C)
m ≈ 0.00144 kg
Therefore, the mass of water in the jug before heating is approximately 0.00144 kg.
To find the mass of water in the jug before heating, we need to use the formula:
Q = mcΔT
Where:
Q is the heat transferred to the water
m is the mass of the water
c is the specific heat capacity of water
ΔT is the change in temperature
Given information:
Specific heat capacity of water, c = 250 J/kg
Initial temperature, T1 = 28°C
Final temperature, T2 = 100°C
First, we need to calculate the heat transferred to the water.
Step 1: Calculate the change in temperature (ΔT)
ΔT = T2 - T1
ΔT = 100°C - 28°C
ΔT = 72°C
Step 2: Calculate the heat transferred to the water (Q)
Q = mcΔT
Since we are given the specific heat capacity, we can rearrange the formula to solve for mass (m):
m = Q / (cΔT)
Step 3: Determine the value of Q
Q is the total energy transferred, which can be calculated using the equation:
Q = P × t
Where:
P is the power
t is the time taken
Given information:
Resistance of the heater, R = 35 ohms
Voltage of the source, V = 220 volts
Time taken to heat the water, t = 4 minutes = 4 × 60 seconds = 240 seconds
Step 4: Calculate the power (P) using the formula:
P = V^2 / R
P = (220 volts)^2 / 35 ohms
P = 48400 / 35
P = 1382.86 watts
Now, we can calculate Q using the formula:
Q = P × t
Q = 1382.86 watts × 240 seconds
Q = 331468.4 joules
Finally, plug in the values of Q, c, and ΔT into the formula to calculate the mass of water (m):
m = Q / (cΔT)
m = 331468.4 joules / (250 J/kg × 72 °C)
m = 4.603176 kg
Therefore, the mass of the water in the jug before heating is approximately 4.60 kg.