A piece of copper weighing 250 is heated to 100 and quickly transferred to a copper calorimeter of mass 15 containing 80 of liquid with an unknow specific heat capacity at a room temperature of 10.caluate the SHC of the liquid.the answer please

Question

A piece of copper weighing 250 is heated to 100 and quickly transferred to a copper calorimeter of mass 15 containing 80 of liquid with an unknow specific heat capacity at a room temperature of 10.caluate the SHC of the liquid.the answer please

Answer 1

Am need of the answer please

Answer 2

Am in need of the answer please

Answer 3

missing the final temperature

Answer 4

To calculate the specific heat capacity (SHC) of the liquid, we can use the principle of conservation of energy.

The amount of heat gained by the liquid can be calculated using the formula:

Q = mcΔT

Where:
Q = Heat gained by the liquid
m = mass of liquid
c = specific heat capacity of the liquid
ΔT = change in temperature

First, let's calculate the amount of heat transferred from the piece of copper to the liquid:

Q1 = mcΔT1

Where:
m = mass of copper = 250 g
c = specific heat capacity of copper (which is known to be 0.39 J/g°C)
ΔT1 = change in temperature of copper

Since the copper is heated to 100°C and the room temperature is 10°C, the change in temperature of the copper is:

ΔT1 = 100 - 10 = 90°C

So, the amount of heat transferred from the copper to the liquid is:

Q1 = (250 g)(0.39 J/g°C)(90°C) = 8775 J

Next, let's calculate the amount of heat gained by the liquid:

Q2 = mlΔT2

Where:
m = mass of liquid = 80 g
l = specific heat capacity of the liquid (which we need to find)
ΔT2 = change in temperature of the liquid

Given that the mass of the copper calorimeter is 15 g, the total mass of the system (liquid + calorimeter) is:

Total mass = mass of liquid + mass of calorimeter = 80 g + 15 g = 95 g

Since the change in temperature of the liquid is from room temperature (10°C) to the final temperature, which is the same as the final temperature of the copper, we can rewrite the equation as:

Q2 = (95 g)(l)(90°C)

We know that the total heat gained by the system is equal to the heat lost by the copper:

Q1 = Q2

So, we have:

8775 J = (95 g)(l)(90°C)

Now, we can solve for the specific heat capacity of the liquid (l):

l = 8775 J / (95 g * 90°C) = 0.104 J/g°C

Therefore, the specific heat capacity of the liquid is 0.104 J/g°C.