Pls help me God will bless you all. A piece of copper ball of mass 20 g at 200 degree Celsius is placed in a 50 g of water at 30 degree Celsius, ignoring heat losses, calculate the final steady temperature of the mixture. Specific heat capacity of water=4.2j/ kg.g

sum of heats gained = zero.

.020*specheatCu*(Tf-200) +50*4.2*(Tf-30)=0
solve for Tf. Watch units

To calculate the final steady temperature of the mixture, we can use the principle of thermal equilibrium. This principle states that when two objects at different temperatures come into contact, heat transfer occurs until they reach a common final temperature.

First, let's find the heat gained or lost by each object.

For the copper ball:
Mass of copper ball = 20 g = 0.02 kg
Initial temperature of copper ball = 200 degree Celsius
Specific heat capacity of copper = ?

Since the specific heat capacity of copper is not provided, we will assume it to be 386 J/kg·K.

Heat gained or lost by the copper ball (Q1) = mass × specific heat capacity × change in temperature
Q1 = 0.02 kg × 386 J/kg·K × (final temperature - 200)

For the water:
Mass of water = 50 g = 0.05 kg
Initial temperature of water = 30 degree Celsius
Specific heat capacity of water = 4.2 J/g·°C = 4200 J/kg·K

Heat gained or lost by the water (Q2) = mass × specific heat capacity × change in temperature
Q2 = 0.05 kg × 4200 J/kg·K × (final temperature - 30)

According to the principle of thermal equilibrium, Q1 = -Q2 (since heat lost by one object is gained by the other).
Therefore, we can equate the two equations:

0.02 × 386 × (final temperature - 200) = -0.05 × 4200 × (final temperature - 30)

Simplifying the equation, we get:

7.72 × final temperature - 7.72 × 200 = -210 × final temperature + 6300

Combining like terms:

217 × final temperature = 6172

Dividing both sides by 217:

final temperature ≈ 28.47 degrees Celsius

Therefore, the final steady temperature of the mixture is approximately 28.47 degrees Celsius.