(C) The heat energy required to raise the temperature of water is given by $Q = m \cdot s \cdot \Delta T$.Here,$m = 2\,kg$,$s = 4200\,J\,kg^{-1}K^{-1}

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(C) The heat energy required to raise the temperature of water is given by $Q = m \cdot s \cdot \Delta T$.Here,$m = 2\,kg$,$s = 4200\,J\,kg^{-1}K^{-1}$,and $\Delta T = 60^{\circ}C - 10^{\circ}C = 50^{\circ}C$.$Q = 2 \times 4200 \times 50 = 420,000\,J$.The effective power delivered by the heater is $P_{eff} = \eta \times P = 0.70 \times 2000\,W = 1400\,W$.The time required is $\Delta t = \frac{Q}{P_{eff}} = \frac{420,000}{1400} = 300\,s$.

Class 11 Physics 10-1.Thermometry, Thermal Expansion and Calorimetry Principle of Calorimetry and Water Equivalent

Medium

$A$ water heater of power $2000\,W$ is used to heat water. The specific heat capacity of water is $4200\,J\,kg^{-1}\,K^{-1}$. The efficiency of the heater is $70\%$. The time required to heat $2\,kg$ of water from $10^{\circ}C$ to $60^{\circ}C$ is $..........s$. (Assume that the specific heat capacity of water remains constant over the temperature range).

JEE MAIN 2023 All JEE MAIN

A
$301$
B
$302$
C
$300$
D
$303$

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10-1.Thermometry, Thermal Expansion and Calorimetry

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Principle of Calorimetry and Water Equivalent

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One kilogram of ice at $0\,^{\circ}C$ is mixed with one kilogram of water at $80\,^{\circ}C$. The temperature of the mixture is ........ $^{\circ}C$ (specific heat of water $= 4200\,J\,kg^{-1}\,K^{-1}$ and latent heat of ice $= 336\,kJ\,kg^{-1}$).

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Ice at $0^oC$ is added to $200\,g$ of water initially at $70^oC$ in a vacuum flask. When $50\,g$ of ice has been added and has all melted,the temperature of the flask and contents is $40^oC$. When a further $80\,g$ of ice has been added and has all melted,the temperature of the whole is $10^oC$. Calculate the specific latent heat of fusion of ice. [Take $S_w = 1\,cal/g^oC$.]

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$A$ $2 \, kg$ block of ice at $-20^{\circ}C$ is added to $5 \, kg$ of water at $20^{\circ}C$. What will be the total mass of water in $kg$? (Specific heat of water = $1 \, kcal/kg/^{\circ}C$,specific heat of ice = $0.5 \, kcal/kg/^{\circ}C$,latent heat of fusion of ice = $80 \, kcal/kg$)

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$A$ $210 \,W$ heater is used to heat $100 \,g$ of water. The time required to raise the temperature of this water from $25^{\circ} C$ to $100^{\circ} C$ is (specific heat capacity of water $= 4200 \,J / kg \cdot ^{\circ} C$). (in $\,s$)

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$A$ liquid of mass $2m$ and specific heat $C$ is heated to a temperature $4T$. Another liquid of mass $m$ and specific heat $2C$ is heated to a temperature $T$. If these two liquids are mixed,the resulting temperature of the mixture is:

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One kilogram of ice at $0\,^{\circ}C$ is mixed with one kilogram of water at $80\,^{\circ}C$. The temperature of the mixture is ........ $^{\circ}C$ (specific heat of water $= 4200\,J\,kg^{-1}\,K^{-1}$ and latent heat of ice $= 336\,kJ\,kg^{-1}$).

$A$ $2 \, kg$ block of ice at $-20^{\circ}C$ is added to $5 \, kg$ of water at $20^{\circ}C$. What will be the total mass of water in $kg$? (Specific heat of water = $1 \, kcal/kg/^{\circ}C$,specific heat of ice = $0.5 \, kcal/kg/^{\circ}C$,latent heat of fusion of ice = $80 \, kcal/kg$)

$A$ $210 \,W$ heater is used to heat $100 \,g$ of water. The time required to raise the temperature of this water from $25^{\circ} C$ to $100^{\circ} C$ is (specific heat capacity of water $= 4200 \,J / kg \cdot ^{\circ} C$). (in $\,s$)

$A$ liquid of mass $2m$ and specific heat $C$ is heated to a temperature $4T$. Another liquid of mass $m$ and specific heat $2C$ is heated to a temperature $T$. If these two liquids are mixed,the resulting temperature of the mixture is:

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