A beaker contains 50 gm water at 20 degree c.How much water having temperature 60 degree C should be added to have an equilibrium temperature of 40 degree c?

C * m (60-40) = C * 50 (40-20)

so
m = 50 grams or .050 Kg

To solve this problem, we can use the principle of heat transfer and assume that there are no heat losses to the surroundings.

Let's denote the mass of water to be added as 'm' grams.

The heat gained by the water can be calculated using the formula:

Heat gained = mass × specific heat capacity × temperature change

For the water already in the beaker, the heat gained will be:

Heat gained by 50g of water = 50g × specific heat capacity of water × (equilibrium temperature - initial temperature)

Similarly, the heat gained by the additional water will be:

Heat gained by m grams of water = m grams × specific heat capacity of water × (equilibrium temperature - temperature of the added water)

According to the principle of heat transfer, the total heat gained by the system should be equal to zero. Therefore, we can set up the equation:

Heat gained by 50g of water + Heat gained by m grams of water = 0

Substituting the respective values, we get:

50g × specific heat capacity of water × (equilibrium temperature - initial temperature) + m grams × specific heat capacity of water × (equilibrium temperature - temperature of the added water) = 0

Substituting the values given in the problem:

50g × 1 cal/g°C × (40°C - 20°C) + m grams × 1 cal/g°C × (40°C - 60°C) = 0

Simplifying the equation gives:

50g × 20°C + m grams × (-20°C) = 0

Now, we can solve for 'm' by rearranging the equation:

m grams × (-20°C) = -50g × 20°C

m grams = (50g × 20°C) / 20°C

m grams = 50g

Therefore, 50 grams of water at 60°C should be added to the beaker to achieve an equilibrium temperature of 40°C.

To find the amount of water needed to reach an equilibrium temperature, we can use the principle of heat transfer. The formula for heat transfer is:

Q = mcΔT

Where:
Q = heat transfer (in calories)
m = mass of the substance (in grams)
c = specific heat capacity of the substance (in calories per gram degree Celsius)
ΔT = change in temperature (in degree Celsius)

In this problem, the beaker of water at 20 degrees Celsius will lose heat (Q1), while the additional water at 60 degrees Celsius will gain heat (Q2). The sum of these two heat transfers should be zero in order to reach equilibrium:

Q1 + Q2 = 0

Let's calculate the heat transfer for each part and solve the equation.

1) For the initial water at 20 degrees Celsius:
Q1 = mcΔT = (50g) * (1 cal/g °C) * (40°C - 20°C) = 1000 calories

2) For the additional water at 60 degrees Celsius:
Q2 = mcΔT = (m g) * (1 cal/g °C) * (40°C - 60°C) = -20m calories

Since both heat transfers need to cancel each other out:
Q1 + Q2 = 0
1000 calories - 20m calories = 0
-20m = -1000
m = 1000 / 20
m = 50 grams

Therefore, 50 grams of water at 60 degrees Celsius need to be added to the beaker in order to reach an equilibrium temperature of 40 degrees Celsius.