Work of 730 J is done by stirring an insulated beaker containing 80 g of water.
What is the change in the temperature of the water?
Please show calculation
To calculate the change in temperature of the water, we can use the equation:
Q = m * c * ΔT
where
Q = heat energy transferred to the water (in joules)
m = mass of the water (in grams)
c = specific heat capacity of water (4.18 J/g°C)
ΔT = change in temperature (in °C)
Given:
Q = 730 J
m = 80 g
c = 4.18 J/g°C
Plugging in these values into the formula, we can solve for ΔT:
730 J = 80 g * 4.18 J/g°C * ΔT
Simplifying the equation:
ΔT = 730 J / (80 g * 4.18 J/g°C)
ΔT ≈ 2.2 °C
Therefore, the change in temperature of the water is approximately 2.2°C.
To calculate the change in temperature of the water, we can use the equation:
Q = m * c * ΔT
Where:
Q = Heat energy (in joules)
m = Mass of water (in grams)
c = Specific heat capacity of water (4.18 J/g°C)
ΔT = Change in temperature (in °C)
Given:
Q = 730 J
m = 80 g
c = 4.18 J/g°C
We need to rearrange the equation to solve for ΔT:
ΔT = Q / (m * c)
Now we can substitute the given values into the equation and solve for ΔT:
ΔT = 730 J / (80 g * 4.18 J/g°C)
ΔT = 730 J / 334.4 J/°C
ΔT ≈ 2.18 °C
Therefore, the change in temperature of the water is approximately 2.18 °C.
The work done becomes the same as heat addition, Q = 730J = 174.5 calories.
Q = C M *(delta T)
C = 1.0 cal/(g*degC)
M = 80 g
Solve for deltaT, the water temperature change, in degrees C
deltaT = 174.5 cal/[(1.0 cal/degC)*(80 g)] = 2.2 deg C