Principle of Calorimetry

For shown situation, calculate the temperature of the common interface.

There are three liquids $A,B$ and $C$. The mass, heat capacity and temperature for liquid $A$ is $\left(m_A,C_A\right.$, and $\left.T_A\right)$, for liquid $B$ is $\left(m_B,C_B\right.$, and $\left.T_B\right)$ and for liquid $C$ is $\left(m_C,C_C\right.$, and $\left.T_C\right)$ respectively. If all the three liquids are mixed, then final temperature would be

A continuous flow water heater (geyser) has an electrical power rating $=2 \mathrm{kW}$ and efficiency of conversion of electrical power into heat $80\%$. If water is flowing through the device at the rate of $100 \mathrm{cc}/\sec$, and the inlet temperature is $10 °\mathrm{C}$, the outlet temperature will be

What will be the final temperature if $50 \mathrm{g}$ of water at $0° \mathrm{C}$ is added to $250 \mathrm{g}$ of water at $90° \mathrm{C}$? 

If the Ropes of masses $M$ & $M^{\text{'}}$ are Replaced by the Massless Strings of Similar Corresponding Lengths. The String attached to masses $m_1 \& m_2$ is cut at the point Just above the mass $m_2$  A Machine which converts Mechanical Energy into Heat Energy is placed $5 \mathrm{m}$ below the (Rope of) Mass $m_2$ before the Ropes are replaced. Efficiency $\left(\eta \%\right)$ of machine is $50\%$. The Heat is supplied to a Container which consists of $0.5 \mathrm{kg}$ of Ice at $-10 °\mathrm{C}$. Find the Final Temp. & composition Pn the container?

$\underset{\left(\mathrm{mass} \mathrm{of} \mathrm{rope} \left(1\right)\right)}{M=10 \mathrm{Kg}}, \underset{\left(\mathrm{length} \mathrm{of} \mathrm{rope} \left(1\right)\right)}{L^{\text{'}}=5 \mathrm{m}}$

$\underset{\left(\mathrm{Area} \mathrm{of} \mathrm{rope} \left(2\right)\right)}{A^{\text{'}}={10}^{-5} {\mathrm{m}}^2}, \underset{\left(\mathrm{foug}\text{'}\mathrm{s} \mathrm{modulues} \left(2\right)\right)}{Y^{\text{'}}={10}^9 \mathrm{N} {\mathrm{m}}^{-2}}$

$m_1=60 \mathrm{kg}, m_2=20 \mathrm{kg}, g=10 \mathrm{m} {\mathrm{s}}^{-2}, \underset{\mathrm{Ice}}{\left(\mathrm{Specific} \mathrm{Heat}\right)}=2100 \mathrm{J} {\mathrm{kg}}^{-1} {}^{\circ }\mathrm{C}^{-1}$

Suppose the masses of the calorimeter, the water in it and the hot object made of copper which is put in the calorimeter are the same. The initial temperature of the calorimeter and water is $30{ }^{°}\mathrm{C}$ and that of the hot object is $60{ }^{°}\mathrm{C}$. The specific heats of copper and water are $0.09 \mathrm{cal}/(\mathrm{gm} °\mathrm{C})$ and $1 \mathrm{cal}/(\mathrm{gm} °\mathrm{C})$ respectively. What will be the final temperature of water?

You are provided some lead shots. How do you find the specific heat of lead?

Initially, a beaker has $100 \mathrm{g}$ of water at temperature $90 °\mathrm{C}$. Later another $600 \mathrm{g}$ of water at temperature $20 °\mathrm{C}$ was poured into the beaker. The temperature, $T\left(\text{in} °\mathrm{C}\right)$ of the water after mixing is

A thermally insulated vessel contains some water at $0°\mathrm{C}$. The vessel is connected to a vacuum pump to pump out water vapour. This results in some water getting frozen. It is given latent heat of vaporization of water at $0°\mathrm{C}=21\times {10}^5 \mathrm{J}/\mathrm{kg}$ and latent heat of freezing of water $=3.36\times {10}^5 \mathrm{J}/\mathrm{kg}$. The maximum percentage amount of water that will be solidified in this manner will be:-

Why the container of calorimeter is made up of copper?

What is calorimeter?

In a certain experiment, liquid helium at an initial temperature of $1 \mathrm{K}$ is to be further cooled by bringing it into contact with a paramagnetic salt which is initially at temperature $T_0$. The heat capacity of liquid helium is expressed as $C_h=a{T}^3\left(a=128\times {10}^{-4}\mathrm{J}·{\mathrm{K}}^{-4}\right)$  and that of salt is expressed as $C_s=b{T}^{-2}\left(b=15\times {10}^{-4} \mathrm{J}.\mathrm{K}\right)$. Assume that the mixture is thermally and mechanically isolated and no work is done during the process. If the final temperature of the system is $0.5 \mathrm{K}$, the initial temperature $T_0$ of the salt is

$1 \mathrm{kg}$ water of specific heat $1 \mathrm{cal} {\mathrm{g}}^{-1}{}^{\circ }\mathrm{C}^{-1}$ is kept in a container at $10°\mathrm{C}$. If $500 \mathrm{g}$ of ice at $0°\mathrm{C}$ is required to cool down the water from $10°\mathrm{C}$ to $0°\mathrm{C}$, then what is the water equivalent of the container? (Latent heat of fusion for ice $L=80 \mathrm{cal} {\mathrm{g}}^{-1}$)

What will be the maximum percentage amount of water that solidifies, if a thermally insulated vessel contains some water at $0°\mathrm{C}$? The vessel is connected to a vacuum pump to pump out water vapour. This results in some water getting frozen. It is given Latent heat of vaporization of water at $0°\mathrm{C}=21\times {10}^5$ $\mathrm{J} {\mathrm{kg}}^{-1}$ and latent heat of freezing of water $=3.36\times {10}^5$ $\mathrm{J} {\mathrm{kg}}^{-1}$. Give answer in the nearest integer.

Two blocks of ice when pressed together join to
form one block. It happens because

When inflated tyre bursts, the air escaps out.

When salt is properly mixed with ice, the melting point of ice?

A circular disc of copper has a symmetrical hole at its centre. The disc is uniformly heated. The diameter of the hole will

Ice in a freezer is at $-7°\mathrm{C}.100 \mathrm{g}$ of this ice is mixed with $200 \mathrm{g}$ of water at $15°\mathrm{C}$. Take the freezing temperature of water to be $0°\mathrm{C}$, the specific heat of ice equal to $2.2 \mathrm{J}/\mathrm{g}°\mathrm{C},$ specific heat of water equal to$4.2 \mathrm{J}/\mathrm{g}°\mathrm{C}$ and the latent heat of ice equal to $335 \mathrm{J}/\mathrm{g}$.
Assuming no loss of heat to the environment, the mass of ice in the final mixture is closest to
 

$50 \mathrm{g}$ of copper is heated to increase its temperature by $10°\mathrm{C}.$ If the same quantity of heat is given to $10 \mathrm{g}$ of water, the rise in temperature is $\_\_\_$ (specific heat of $Cu=420$ $\mathrm{J}·{\mathrm{kg}}^{-1}·{{}^{\circ }\mathrm{C}}^{-1}$ and specific heat of water is $\left.4200 \mathrm{J}.{\mathrm{kg}}^{-1}·{{}^{\circ }\mathrm{C}}^{-1}\right)$