1. (a) 500g of copper is heated from 30 degree celcius - 120 degree celcius?. (b) if the same quantity of heat is supplied to 300g of water initial at 30 degree celcius,what will be the final temperature.
Sorry. I could only give direct answer not full descriptive solution
Answer: approx. 43.77 °C.
To calculate the temperature change in copper and the final temperature of water, we can use the formula:
Q = mcΔT
Where:
Q = heat energy
m = mass
c = specific heat capacity
ΔT = change in temperature
For the copper:
(a)
Given:
Initial temperature (Ti) = 30°C
Final temperature (Tf) = 120°C
Mass (m) = 500g
Specific heat capacity (c) = 0.385 J/g°C (for copper)
Using the formula, we can calculate the heat energy (Q1) absorbed by the copper:
Q1 = mcΔT1
= (500g)(0.385 J/g°C)(120°C - 30°C)
= (500g)(0.385 J/g°C)(90°C)
≈ 17,325 J
For the water:
(b)
Given:
Initial temperature (Ti) = 30°C
Mass (m) = 300g
Specific heat capacity (c) = 4.184 J/g°C (for water)
Using the formula, we can calculate the change in temperature (ΔT2) for the water:
Q2 = mcΔT2
= (300g)(4.184 J/g°C)(ΔT2)
Since the same quantity of heat is supplied to both the copper and the water, Q1 = Q2. Therefore, we can equate the two equations:
Q1 = Q2
17,325 J = (300g)(4.184 J/g°C)(ΔT2)
Simplifying the equation:
ΔT2 = 17,325 J / [(300g)(4.184 J/g°C)]
≈ 14.9°C
To find the final temperature (Tf) of the water, we add ΔT2 to the initial temperature (Ti):
Tf = Ti + ΔT2
= 30°C + 14.9°C
≈ 44.9°C
Therefore, the final temperature of the water will be approximately 44.9°C.
To find the change in temperature for a substance when heat is supplied, we can use the formula:
Q = mcΔT
Where:
Q = heat energy absorbed or released (in joules)
m = mass of the substance
c = specific heat capacity of the substance
ΔT = change in temperature
(a) Let's find the change in temperature for 500g of copper when heated from 30 degrees Celsius to 120 degrees Celsius. The specific heat capacity of copper is approximately 0.39 J/g°C.
Q = mcΔT
Since ΔT = T2 - T1, where T1 is the initial temperature and T2 is the final temperature:
ΔT = 120°C - 30°C = 90°C
Substituting the values:
Q = (500g) × (0.39 J/g°C) × (90°C)
Q = 17550 J
Therefore, 500g of copper heated from 30°C to 120°C will absorb 17550 joules of heat.
(b) Now let's find the final temperature when the same quantity of heat is supplied to 300g of water initially at 30 degrees Celsius.
Again, using the formula Q = mcΔT, but this time for water. The specific heat capacity of water is approximately 4.18 J/g°C.
Q = mcΔT
We know that Q = 17550 J, m = 300g, and T1 = 30°C. Let's assume the final temperature is T2.
17550 J = (300g) × (4.18 J/g°C) × (T2 - 30°C)
Solving for T2:
17550 J = 1254g°C × (T2 - 30°C)
T2 - 30°C = 17550 J / 1254 g°C
T2 - 30°C ≈ 14°C
Adding 30°C to both sides of the equation:
T2 ≈ 14°C + 30°C
T2 ≈ 44°C
Therefore, when the same quantity of heat is supplied to 300g of water initially at 30°C, the final temperature of the water will be approximately 44°C.