Class 11Physics10-1.Thermometry, Thermal Expansion and CalorimetryPrinciple of Calorimetry and Water Equivalent
MediumThree copper blocks of masses $M_1, M_2$ and $M_3$ $kg$ respectively are brought into thermal contact until they reach equilibrium. Before contact,they were at temperatures $T_1, T_2, T_3$ $(T_1 > T_3)$. Assuming there is no heat loss to the surroundings,the equilibrium temperature $T$ is ($s$ is the specific heat of copper).
- A$T = \frac{T_1 + T_2 + T_3}{3}$
- B$T = \frac{M_1 T_1 + M_2 T_2 + M_3 T_3}{M_1 + M_2 + M_3}$
- C$T = \frac{M_1 T_1 + M_2 T_2 + M_3 T_3}{3(M_1 + M_2 + M_3)}$
- D$T = \frac{M_1 T_1 s + M_2 T_2 s + M_3 T_3 s}{3(M_1 + M_2 + M_3)}$
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Similar Questions
Three containers $C_{1}, C_{2}$ and $C_{3}$ have water at different temperatures. The table below shows the final temperature $T$ when different amounts of water (given in litres) are taken from each container and mixed (assume no loss of heat during the process).
$C_{1}$ $C_{2}$ $C_{3}$ $T$ $1 \ l$ $2 \ l$ $-$ $60^{\circ} C$ $-$ $1 \ l$ $2 \ l$ $30^{\circ} C$ $2 \ l$ $-$ $1 \ l$ $60^{\circ} C$ $1 \ l$ $1 \ l$ $1 \ l$ $\theta$
The value of $\theta$ (in $^{\circ} C$ to the nearest integer) is
| $C_{1}$ | $C_{2}$ | $C_{3}$ | $T$ |
| $1 \ l$ | $2 \ l$ | $-$ | $60^{\circ} C$ |
| $-$ | $1 \ l$ | $2 \ l$ | $30^{\circ} C$ |
| $2 \ l$ | $-$ | $1 \ l$ | $60^{\circ} C$ |
| $1 \ l$ | $1 \ l$ | $1 \ l$ | $\theta$ |
The value of $\theta$ (in $^{\circ} C$ to the nearest integer) is
DifficultJEE MAIN 2020
View Solution$A$ piece of copper of mass $250 \ g$ at $500^{\circ} C$ is put inside a calorimeter of water equivalent $50 \ g$ containing $200 \ g$ of water at $20^{\circ} C$. At thermal equilibrium,the temperature of the mixture is $60^{\circ} C$. The specific heat of copper (in $J / kg \cdot ^{\circ} C$) is approximately: [Specific heat of water $= 4200 \ J / kg \cdot ^{\circ} C$]
Medium
View SolutionThe time taken for a calorimeter containing $75 \ g$ of water at $62^{\circ} C$ to cool to $58^{\circ} C$ is $9 \ minutes$. When the calorimeter contains $105 \ g$ of water,it takes $12 \ minutes$ to cool from $62^{\circ} C$ to $58^{\circ} C$. The water equivalent of the calorimeter is $.........$ (in $g$)
$A$ $10 \ W$ electric heater is used to heat a container filled with $0.5 \ kg$ of water. It is found that the temperature of the water and the container rises by $3 \ K$ in $15 \ min$. The container is then emptied, dried, and filled with $2 \ kg$ of oil. The same heater now raises the temperature of the container-oil system by $2 \ K$ in $20 \ min$. Assuming that there is no heat loss in the process and the specific heat of water is $4200 \ J \ kg^{-1} \ K^{-1}$, the specific heat of oil in the same unit is equal to:
DifficultWBJEE 2014
View SolutionSteam at $100^{\circ}C$ is added slowly to $1400 \, \text{g}$ of water at $16^{\circ}C$ until the temperature of the water is raised to $80^{\circ}C$. The mass of steam required to do this is $(L_V = 540 \, \text{cal/g})$ ........... $\text{g}$
Difficult
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