an electric heater raises the temperature of 120g of water in a thin light vessel through 10K in 2mins, when placed in 70g of water contained in a metal vessel of mass 0.55kg the temperature rises through 9K in the same time. Calculate from the above:
a)the heat supplied in 2mins
b)the power of the heater
c)the heat supplied to the 70g of water
d)the heat supplied to the metal vessel
e)the heat capacity of the vessel
f) the specific heat capacity of its material (specific heat capacity of water = 4200)
i nead d answer please
show d workings please
To solve this problem, we need to use the formula:
Q = mcΔθ
Where:
Q is the heat supplied
m is the mass of the substance
c is the specific heat capacity of the substance
Δθ is the change in temperature
a) The heat supplied in 2 minutes:
For the 120g of water:
Q1 = m1 * c * Δθ1
= 120g * 4200 J/kg°C * 10K
= 5,040,000 J
For the 70g of water:
Q2 = m2 * c * Δθ2
= 70g * 4200 J/kg°C * 9K
= 2,646,000 J
Therefore, the total heat supplied in 2 minutes is:
Q = Q1 + Q2
= 5,040,000 J + 2,646,000 J
= 7,686,000 J
b) The power of the heater:
Power (P) is the rate at which energy is transferred, and it can be calculated using the formula:
P = Q / t
Where:
P is the power
Q is the heat supplied
t is the time
From the given information, the time (t) is 2 minutes, which is equal to 120 seconds.
P = 7,686,000 J / 120 s
= 64,050 W
Therefore, the power of the heater is 64,050 Watts.
c) The heat supplied to the 70g of water:
We already calculated the heat supplied to the 70g of water in part a:
Q2 = 2,646,000 J
Therefore, the heat supplied to the 70g of water is 2,646,000 Joules.
d) The heat supplied to the metal vessel:
To calculate the heat supplied to the metal vessel, we need to subtract the heat supplied to the water (Q2) from the total heat supplied (Q).
Heat supplied to the metal vessel = Q - Q2
= 7,686,000 J - 2,646,000 J
= 5,040,000 J
Therefore, the heat supplied to the metal vessel is 5,040,000 Joules.
e) The heat capacity of the vessel:
The heat capacity of an object is defined as the amount of heat energy required to raise its temperature by 1 degree Celsius.
Heat capacity (C) can be calculated using the formula:
C = Q / Δθ
Where:
C is the heat capacity
Q is the heat supplied
Δθ is the change in temperature
For the metal vessel:
C = 5,040,000 J / 9K
= 560,000 J/K
Therefore, the heat capacity of the metal vessel is 560,000 Joules per Kelvin (J/K).
f) The specific heat capacity of its material:
The specific heat capacity is defined as the amount of heat energy required to raise the temperature of 1 kilogram of a substance by 1 degree Celsius.
We have already calculated the heat capacity of the vessel (C). Now, we can use the formula:
C = mc
Where:
C is the heat capacity
m is the mass of the substance
c is the specific heat capacity of the substance
For the metal vessel, the mass (m) is 0.55kg.
C = 0.55kg * c
Since we want to find the specific heat capacity of the material, we rearrange the equation as:
c = C / m
= 560,000 J/K / 0.55kg
= 1,018,181.82 J/kg°C
Therefore, the specific heat capacity of the material is approximately 1,018,181.82 Joules per kilogram per degree Celsius (J/kg°C).