$540\, g$ of ice at $0^oC$ is mixed with $540\, g$ of water at $80^oC$. The final temperature of the mixture is
$540\, g$ of ice at $0^oC$ is mixed with $540\, g$ of water at $80^oC$. The final temperature of the mixture is
- A
$0°C$
- B
$40°C$
- C
$80°C$
- D
Less than $0°C$
Similar Questions
A bullet of mass $10 \,g$ moving with a speed of $20 \,m / s$ hits an ice block of mass $990 \,g$ kept on a frictionless floor and gets stuck in it. How much ice will melt if $50 \%$ of the lost KE goes to ice is .......... $g$ (initial temperature of the ice block and bullet $=0^{\circ} C$ )
A bullet of mass $10 \,g$ moving with a speed of $20 \,m / s$ hits an ice block of mass $990 \,g$ kept on a frictionless floor and gets stuck in it. How much ice will melt if $50 \%$ of the lost KE goes to ice is .......... $g$ (initial temperature of the ice block and bullet $=0^{\circ} C$ )
A $100 \,g$ of iron nail is hit by a $1.5\, kg$ hammer striking at a velocity of $60\, ms ^{-1}$. $............^{\circ}C$ will be the rise in the temperature of the nail if one fourth of energy of the hammer goes into heating the nail .
[Specific heat capacity of iron $=0.42\, Jg ^{-1}{ }\,^{\circ}C ^{-1}$ ]
A $100 \,g$ of iron nail is hit by a $1.5\, kg$ hammer striking at a velocity of $60\, ms ^{-1}$. $............^{\circ}C$ will be the rise in the temperature of the nail if one fourth of energy of the hammer goes into heating the nail .
[Specific heat capacity of iron $=0.42\, Jg ^{-1}{ }\,^{\circ}C ^{-1}$ ]
- [JEE MAIN 2022]
The specific heat of alcohol is about half that of water. Suppose you have identical masses of alcohol and water. The alcohol is initially at temperature $T_A$ . The water is initially at a different temperature $T_W$ . Now the two fluids are mixed in the same container and allowed to come into thermal equilibrium, with no loss of heat to the surroundings. The final temperature of the mixture will be
The specific heat of alcohol is about half that of water. Suppose you have identical masses of alcohol and water. The alcohol is initially at temperature $T_A$ . The water is initially at a different temperature $T_W$ . Now the two fluids are mixed in the same container and allowed to come into thermal equilibrium, with no loss of heat to the surroundings. The final temperature of the mixture will be
The specific heat of a metal at low temperatures varies according to $S = aT^3$ where a is a constant and $T$ is the absolute temperature. The heat energy needed to raise unit mass of the metal from $T = 1 K$ to $T = 2 K$ is
The specific heat of a metal at low temperatures varies according to $S = aT^3$ where a is a constant and $T$ is the absolute temperature. The heat energy needed to raise unit mass of the metal from $T = 1 K$ to $T = 2 K$ is
Find the quantity of heat required to convert $40\; gm$ of ice at $-20^{\circ} C$ into water at $20^{\circ} C$. Given $L _{\text {ice }}$ $=0.336 \times 10^6 J / kg$.
specific heat of ice $=2100 \;J / kg - K$ sp heat of water= $4200\; J / kg - K$
Find the quantity of heat required to convert $40\; gm$ of ice at $-20^{\circ} C$ into water at $20^{\circ} C$. Given $L _{\text {ice }}$ $=0.336 \times 10^6 J / kg$.
specific heat of ice $=2100 \;J / kg - K$ sp heat of water= $4200\; J / kg - K$