Question

5. A coil of resistance wire is immersed in liquid in a calorimeter of a total heat capacity 950JK. If the temperature rises from 9°C to 29°C in 5 minutes when a steady current of 4A is passed, find: (i) The resistance of the coil​

Answer 1

Answer: the resistance of the coil is 300 ohms.

Step-by-step explanation:

We can use the formula for heat generated by a resistor to find the power generated by the coil:

P = I^2 * R

where P is power, I is current, and R is resistance.

The power generated by the coil will be equal to the heat absorbed by the liquid in the calorimeter:

P = Q/t

where Q is the heat absorbed, and t is the time.

We can use the formula for heat absorbed by a substance to find the heat absorbed by the liquid:

Q = C * m * ΔT

where Q is the heat absorbed, C is the specific heat capacity of the liquid, m is the mass of the liquid, and ΔT is the change in temperature.

We are given the total heat capacity of the calorimeter, which is equal to the sum of the heat capacities of the liquid and the calorimeter:

C_total = C_liquid + C_calorimeter

We can rearrange this equation to solve for the specific heat capacity of the liquid:

C_liquid = C_total - C_calorimeter

Plugging in the given values, we get:

C_liquid = 950 J/K - unknown

We need to find the unknown value of C_calorimeter in order to solve for the specific heat capacity of the liquid. We can do this by using the formula for heat absorbed:

Q = C_total * ΔT

where Q is the heat absorbed by the calorimeter and the liquid, and ΔT is the change in temperature of the calorimeter and the liquid.

Plugging in the given values, we get:

Q = 950 J/K * (29°C - 9°C) = 19000 J

The power generated by the coil is:

P = I^2 * R = 4^2 * R = 16R

The heat absorbed by the liquid is:

Q = P * t = 16R * 300 s = 4800R J

Setting these two equations equal to each other, we get:

16R = 4800R

Dividing both sides by 16, we get:

R = 300 ohms

Therefore, the resistance of the coil is 300 ohms.

Answer 2

0.495 Ω.

Step-by-step explanation:

To solve this problem, we will use the formula for heat:

Q = mcΔT

Where Q is the heat transferred, m is the mass of the liquid, c is the specific heat capacity of the liquid, and ΔT is the change in temperature.

Since the calorimeter has a total heat capacity of 950 J/K, we can write:

Q = 950ΔT

where ΔT is the temperature change of the calorimeter.

The heat generated by the coil can be calculated using the formula:

Q = I²Rt

where I is the current, R is the resistance of the coil, and t is the time the current is passed.

Substituting the given values, we get:

Q = (4A)²R(5min*60s/min) = 960Rs

where Rs is the heat generated per unit resistance.

Since the heat generated by the coil is transferred to the calorimeter, we can equate the two expressions for Q:

960Rs = 950ΔT

Solving for R, we get:

R = (950ΔT) / (960s)

Substituting the given values, we get:

R = (950J/K * (29°C - 9°C)) / (960s * 4²)

R = 0.495 Ω

Therefore, the resistance of the coil is 0.495 Ω.