2,kg of metal at 100,^{circ}C is cooled by 1, | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) According to the principle of calorimetry,heat lost by the metal equals heat gained by the water.Let $m_1 = 2\,kg$ be the mass of the metal and $s

Vedclass · Accepted Answer

$50\,^{\circ}C$

Vedclass · Answer

(A) According to the principle of calorimetry,heat lost by the metal equals heat gained by the water.Let $m_1 = 2\,kg$ be the mass of the metal and $s_1$ be its specific heat capacity.Let $m_2 = 1\,kg$ be the mass of the water and $s_2$ be its specific heat capacity.Given $s_1 = \frac{1}{2} s_2$.Let $	heta$ be the final equilibrium temperature.Heat lost by metal $= m_1 s_1 (100 - 	heta) = 2 	imes (\frac{s_2}{2}) 	imes (100 - 	heta) = s_2 (100 - 	heta)$.Heat gained by water $= m_2 s_2 (	heta - 0) = 1 	imes s_2 	imes 	heta = s_2 	heta$.Equating the two: $s_2 (100 - 	heta) = s_2 	heta$.$100 - 	heta = 	heta$.$2	heta = 100$.$	heta = 50\,^{\circ}C$.

$2\,kg$ of metal at $100\,^{\circ}C$ is cooled by $1\,kg$ of water at $0\,^{\circ}C$. If the specific heat capacity of the metal is $\frac{1}{2}$ of the specific heat capacity of water,the final temperature of the mixture would be:

Similar Questions

$A$ piece of metal of mass $5 \ kg$ and temperature $-50 \ ^\circ C$ is dropped in $20 \ kg$ water at $52 \ ^\circ C$ temperature. In thermal equilibrium,the temperature of water decreases by $2 \ ^\circ C$. Find the specific heat of the metal in $Cal/g \ ^\circ C$.

When $M_1$ gram of ice at $-10\,^{\circ}C$ (specific heat $= 0.5\, cal\, g^{-1}\,^{\circ}C^{-1}$) is added to $M_2$ gram of water at $50\,^{\circ}C$,finally no ice is left and the water is at $0\,^{\circ}C$. The value of latent heat of ice,in $cal\, g^{-1}$ is

We have half a bucket $(6\,L)$ of water at $20\,^oC$. If we want water at $40\,^oC$,how much steam at $100\,^oC$ should be added to it?

Time taken by a $836\; W$ heater to heat $1\; litre$ of water from $10^{\circ} C$ to $40^{\circ} C$ is (in $; s$)

$100 \text{ g}$ of ice at $0^{\circ}C$ is mixed with $100 \text{ g}$ of water at $100^{\circ}C$. The final temperature of the mixture is. [Take,$L_f = 3.36 \times 10^5 \text{ J kg}^{-1}$ and $S_w = 4.2 \times 10^3 \text{ J kg}^{-1} \text{ K}^{-1}$] (in $^{\circ}C$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module