20 gm of ice and 20 gm of hot water are mixed. When the ice is melted the temperature of the mixture was found to be 0'C, the temperature of hot taken should be (L=80cal/gm)-

To find the temperature of the hot water before mixing, we'll use the concept of heat transfer.

First, let's calculate the heat absorbed by the ice when it melts. The heat absorbed is given by the equation:

Q_ice = m_ice * L_f

where Q_ice is the heat absorbed, m_ice is the mass of the ice, and L_f is the latent heat of fusion of water (which is 80 cal/gm).

Given that the mass of the ice is 20 gm, we can calculate Q_ice as follows:

Q_ice = 20 gm * 80 cal/gm = 1600 cal

Since the heat absorbed by the ice comes from the hot water, we can write:

Q_hot = Q_ice

where Q_hot is the heat released by the hot water.

Now, let's calculate the heat released by the hot water using the equation:

Q_hot = m_hot * C * ΔT

where m_hot is the mass of the hot water, C is the specific heat capacity of water (which is approximately 1 cal/gm°C), and ΔT is the change in temperature of the hot water.

Given that the final temperature of the mixture is 0°C, we can write:

ΔT = 0°C - T_hot

where T_hot is the initial temperature of the hot water.

Substituting this into the equation for Q_hot:

Q_hot = m_hot * C * (0°C - T_hot)

Now we can equate Q_hot with Q_ice:

m_hot * C * (0°C - T_hot) = Q_ice

Plugging in the values we know:

m_hot * 1 cal/gm°C * (0°C - T_hot) = 1600 cal

Simplifying the equation:

- m_hot * T_hot = 1600 cal

We're given that the mass of the hot water is also 20 gm:

- 20 gm * T_hot = 1600 cal

Now we can solve for T_hot:

T_hot = 1600 cal / 20 gm
T_hot = 80°C

Therefore, the temperature of the hot water before mixing with the ice should be 80°C.