366367
\(1\;g\) of a steam at \(100^\circ C\) melts how much ice at \(0^\circ C\) (Latent heat of ice \( = 80\,cal/gm\) and latent heat of steam \( = 540\,cal/gm)\)
1 \(2\,gm\)
2 \(1\,gm\)
3 \(8\,gm\)
4 \(4\,gm\)
Explanation:
Suppose \(m\) \(gm\) ice melted, then heat required for its melting \( = mL = m \times 80\,cal\) Heat available with steam for being condensed and then brought to \(0^\circ C\) \( = 1 \times 540 + 1 \times 1 \times (100 - 0) = 640\,cal\) \( \Rightarrow {\text{ Heat lost }} = {\text{ Heat taken }}\) \( \Rightarrow 640 = m \times 80\) \( \Rightarrow m = 8gm\) Short cut : You can remember that amount of steam \(\left(m^{\prime}\right)\) at \(100^{\circ} \mathrm{C}\) required to melt \(m g m\) ice at \(0^{\circ} \mathrm{C}\) is \(m^{\prime}=\dfrac{m}{8} .\) Here, \(m=8 \times m^{\prime}=8 \times 1=8 g m\)
PHXI11:THERMAL PROPERTIES OF MATTER
366368
Heat is supplied to a certain homogeneous sample of matter, at a uniform rate. Its temperature is plotted aganist time, as shown. Which of the following conclusions can be drawn
1 Its specific heat capacity is greater in the liquid state than in the solid state
2 Its specific heat capacity is greater in the solid state than in the liquid state
3 Its latent heat of vaporisation is smaller than its latent of fusion
4 Both (1) and (3) are correct
Explanation:
The horizontal parts of the curve, where the system absorbs heat at constant temperature must represent changes of state. Latent heats are proportional to lengths of the horizontal parts. In the sloping parts, specific heat capacity is inversely proportional to the slopes.
PHXI11:THERMAL PROPERTIES OF MATTER
366369
\(1\,kg\) of ice at \({-10^{\circ} {C}}\) is mixed with \(4.4{\rm{ }}kg\) of water at \({30^{\circ} {C}}\). The final temperature of mixture is (specific heat of ice is \({2100 {~J} / {kg} / {k}}\) )
1 \({2.3^{\circ} {C}}\)
2 \({4.4^{\circ} {C}}\)
3 \({5.3^{\circ} {C}}\)
4 \({8.7^{\circ} {C}}\)
Explanation:
Let temperature of mixture be \({\theta}\), then heat lost \({=}\) heat gained \(\begin{aligned}4.4 \times 1 \times & (30=\theta)=1 \times 0.5 \times(10) \\+ & 1 \times 80+1 \times 1 \times(\theta-0)\end{aligned}\) \({\Rightarrow \theta=8.7^{\circ} {C}}\)
PHXI11:THERMAL PROPERTIES OF MATTER
366370
\(50\;g\) of ice at \( - 5^\circ C\) is added to \(200\;g\) of water at \(40^\circ C\) in a calorimeter. The water equivalent of calorimeter is \(50\;g\) of water. The temperature of the mixture is
1 \(9.58^\circ C\)
2 \(12.58^\circ C\)
3 \(19.58^\circ C\)
4 \(29.58^\circ C\)
Explanation:
As water equivalent of calorimeter is \(50\,g\) of water the total water present in the system is \(250\,g\). Heat absorbed by ice to become water at \(0^\circ C\) is \(Q{\rm{ }} = 50 \times 0.5 \times 5 + 50 \times 80 = 4125\,cal\) The decrease in temperature of water is \(4125 = 250 \times 1 \times \Delta T\) \( \Rightarrow \Delta T = 165^\circ C\) Now we have \(50\;g\) of water at \(0^\circ C\) and \(250\;g\) of water at \(23.5^\circ C\). The temperature of the mixture is \(50 \times 1(T - 0) + 250 \times 1 \times (T - 235) = 0\) \( \Rightarrow T = 19.58^\circ C\)
366367
\(1\;g\) of a steam at \(100^\circ C\) melts how much ice at \(0^\circ C\) (Latent heat of ice \( = 80\,cal/gm\) and latent heat of steam \( = 540\,cal/gm)\)
1 \(2\,gm\)
2 \(1\,gm\)
3 \(8\,gm\)
4 \(4\,gm\)
Explanation:
Suppose \(m\) \(gm\) ice melted, then heat required for its melting \( = mL = m \times 80\,cal\) Heat available with steam for being condensed and then brought to \(0^\circ C\) \( = 1 \times 540 + 1 \times 1 \times (100 - 0) = 640\,cal\) \( \Rightarrow {\text{ Heat lost }} = {\text{ Heat taken }}\) \( \Rightarrow 640 = m \times 80\) \( \Rightarrow m = 8gm\) Short cut : You can remember that amount of steam \(\left(m^{\prime}\right)\) at \(100^{\circ} \mathrm{C}\) required to melt \(m g m\) ice at \(0^{\circ} \mathrm{C}\) is \(m^{\prime}=\dfrac{m}{8} .\) Here, \(m=8 \times m^{\prime}=8 \times 1=8 g m\)
PHXI11:THERMAL PROPERTIES OF MATTER
366368
Heat is supplied to a certain homogeneous sample of matter, at a uniform rate. Its temperature is plotted aganist time, as shown. Which of the following conclusions can be drawn
1 Its specific heat capacity is greater in the liquid state than in the solid state
2 Its specific heat capacity is greater in the solid state than in the liquid state
3 Its latent heat of vaporisation is smaller than its latent of fusion
4 Both (1) and (3) are correct
Explanation:
The horizontal parts of the curve, where the system absorbs heat at constant temperature must represent changes of state. Latent heats are proportional to lengths of the horizontal parts. In the sloping parts, specific heat capacity is inversely proportional to the slopes.
PHXI11:THERMAL PROPERTIES OF MATTER
366369
\(1\,kg\) of ice at \({-10^{\circ} {C}}\) is mixed with \(4.4{\rm{ }}kg\) of water at \({30^{\circ} {C}}\). The final temperature of mixture is (specific heat of ice is \({2100 {~J} / {kg} / {k}}\) )
1 \({2.3^{\circ} {C}}\)
2 \({4.4^{\circ} {C}}\)
3 \({5.3^{\circ} {C}}\)
4 \({8.7^{\circ} {C}}\)
Explanation:
Let temperature of mixture be \({\theta}\), then heat lost \({=}\) heat gained \(\begin{aligned}4.4 \times 1 \times & (30=\theta)=1 \times 0.5 \times(10) \\+ & 1 \times 80+1 \times 1 \times(\theta-0)\end{aligned}\) \({\Rightarrow \theta=8.7^{\circ} {C}}\)
PHXI11:THERMAL PROPERTIES OF MATTER
366370
\(50\;g\) of ice at \( - 5^\circ C\) is added to \(200\;g\) of water at \(40^\circ C\) in a calorimeter. The water equivalent of calorimeter is \(50\;g\) of water. The temperature of the mixture is
1 \(9.58^\circ C\)
2 \(12.58^\circ C\)
3 \(19.58^\circ C\)
4 \(29.58^\circ C\)
Explanation:
As water equivalent of calorimeter is \(50\,g\) of water the total water present in the system is \(250\,g\). Heat absorbed by ice to become water at \(0^\circ C\) is \(Q{\rm{ }} = 50 \times 0.5 \times 5 + 50 \times 80 = 4125\,cal\) The decrease in temperature of water is \(4125 = 250 \times 1 \times \Delta T\) \( \Rightarrow \Delta T = 165^\circ C\) Now we have \(50\;g\) of water at \(0^\circ C\) and \(250\;g\) of water at \(23.5^\circ C\). The temperature of the mixture is \(50 \times 1(T - 0) + 250 \times 1 \times (T - 235) = 0\) \( \Rightarrow T = 19.58^\circ C\)
366367
\(1\;g\) of a steam at \(100^\circ C\) melts how much ice at \(0^\circ C\) (Latent heat of ice \( = 80\,cal/gm\) and latent heat of steam \( = 540\,cal/gm)\)
1 \(2\,gm\)
2 \(1\,gm\)
3 \(8\,gm\)
4 \(4\,gm\)
Explanation:
Suppose \(m\) \(gm\) ice melted, then heat required for its melting \( = mL = m \times 80\,cal\) Heat available with steam for being condensed and then brought to \(0^\circ C\) \( = 1 \times 540 + 1 \times 1 \times (100 - 0) = 640\,cal\) \( \Rightarrow {\text{ Heat lost }} = {\text{ Heat taken }}\) \( \Rightarrow 640 = m \times 80\) \( \Rightarrow m = 8gm\) Short cut : You can remember that amount of steam \(\left(m^{\prime}\right)\) at \(100^{\circ} \mathrm{C}\) required to melt \(m g m\) ice at \(0^{\circ} \mathrm{C}\) is \(m^{\prime}=\dfrac{m}{8} .\) Here, \(m=8 \times m^{\prime}=8 \times 1=8 g m\)
PHXI11:THERMAL PROPERTIES OF MATTER
366368
Heat is supplied to a certain homogeneous sample of matter, at a uniform rate. Its temperature is plotted aganist time, as shown. Which of the following conclusions can be drawn
1 Its specific heat capacity is greater in the liquid state than in the solid state
2 Its specific heat capacity is greater in the solid state than in the liquid state
3 Its latent heat of vaporisation is smaller than its latent of fusion
4 Both (1) and (3) are correct
Explanation:
The horizontal parts of the curve, where the system absorbs heat at constant temperature must represent changes of state. Latent heats are proportional to lengths of the horizontal parts. In the sloping parts, specific heat capacity is inversely proportional to the slopes.
PHXI11:THERMAL PROPERTIES OF MATTER
366369
\(1\,kg\) of ice at \({-10^{\circ} {C}}\) is mixed with \(4.4{\rm{ }}kg\) of water at \({30^{\circ} {C}}\). The final temperature of mixture is (specific heat of ice is \({2100 {~J} / {kg} / {k}}\) )
1 \({2.3^{\circ} {C}}\)
2 \({4.4^{\circ} {C}}\)
3 \({5.3^{\circ} {C}}\)
4 \({8.7^{\circ} {C}}\)
Explanation:
Let temperature of mixture be \({\theta}\), then heat lost \({=}\) heat gained \(\begin{aligned}4.4 \times 1 \times & (30=\theta)=1 \times 0.5 \times(10) \\+ & 1 \times 80+1 \times 1 \times(\theta-0)\end{aligned}\) \({\Rightarrow \theta=8.7^{\circ} {C}}\)
PHXI11:THERMAL PROPERTIES OF MATTER
366370
\(50\;g\) of ice at \( - 5^\circ C\) is added to \(200\;g\) of water at \(40^\circ C\) in a calorimeter. The water equivalent of calorimeter is \(50\;g\) of water. The temperature of the mixture is
1 \(9.58^\circ C\)
2 \(12.58^\circ C\)
3 \(19.58^\circ C\)
4 \(29.58^\circ C\)
Explanation:
As water equivalent of calorimeter is \(50\,g\) of water the total water present in the system is \(250\,g\). Heat absorbed by ice to become water at \(0^\circ C\) is \(Q{\rm{ }} = 50 \times 0.5 \times 5 + 50 \times 80 = 4125\,cal\) The decrease in temperature of water is \(4125 = 250 \times 1 \times \Delta T\) \( \Rightarrow \Delta T = 165^\circ C\) Now we have \(50\;g\) of water at \(0^\circ C\) and \(250\;g\) of water at \(23.5^\circ C\). The temperature of the mixture is \(50 \times 1(T - 0) + 250 \times 1 \times (T - 235) = 0\) \( \Rightarrow T = 19.58^\circ C\)
366367
\(1\;g\) of a steam at \(100^\circ C\) melts how much ice at \(0^\circ C\) (Latent heat of ice \( = 80\,cal/gm\) and latent heat of steam \( = 540\,cal/gm)\)
1 \(2\,gm\)
2 \(1\,gm\)
3 \(8\,gm\)
4 \(4\,gm\)
Explanation:
Suppose \(m\) \(gm\) ice melted, then heat required for its melting \( = mL = m \times 80\,cal\) Heat available with steam for being condensed and then brought to \(0^\circ C\) \( = 1 \times 540 + 1 \times 1 \times (100 - 0) = 640\,cal\) \( \Rightarrow {\text{ Heat lost }} = {\text{ Heat taken }}\) \( \Rightarrow 640 = m \times 80\) \( \Rightarrow m = 8gm\) Short cut : You can remember that amount of steam \(\left(m^{\prime}\right)\) at \(100^{\circ} \mathrm{C}\) required to melt \(m g m\) ice at \(0^{\circ} \mathrm{C}\) is \(m^{\prime}=\dfrac{m}{8} .\) Here, \(m=8 \times m^{\prime}=8 \times 1=8 g m\)
PHXI11:THERMAL PROPERTIES OF MATTER
366368
Heat is supplied to a certain homogeneous sample of matter, at a uniform rate. Its temperature is plotted aganist time, as shown. Which of the following conclusions can be drawn
1 Its specific heat capacity is greater in the liquid state than in the solid state
2 Its specific heat capacity is greater in the solid state than in the liquid state
3 Its latent heat of vaporisation is smaller than its latent of fusion
4 Both (1) and (3) are correct
Explanation:
The horizontal parts of the curve, where the system absorbs heat at constant temperature must represent changes of state. Latent heats are proportional to lengths of the horizontal parts. In the sloping parts, specific heat capacity is inversely proportional to the slopes.
PHXI11:THERMAL PROPERTIES OF MATTER
366369
\(1\,kg\) of ice at \({-10^{\circ} {C}}\) is mixed with \(4.4{\rm{ }}kg\) of water at \({30^{\circ} {C}}\). The final temperature of mixture is (specific heat of ice is \({2100 {~J} / {kg} / {k}}\) )
1 \({2.3^{\circ} {C}}\)
2 \({4.4^{\circ} {C}}\)
3 \({5.3^{\circ} {C}}\)
4 \({8.7^{\circ} {C}}\)
Explanation:
Let temperature of mixture be \({\theta}\), then heat lost \({=}\) heat gained \(\begin{aligned}4.4 \times 1 \times & (30=\theta)=1 \times 0.5 \times(10) \\+ & 1 \times 80+1 \times 1 \times(\theta-0)\end{aligned}\) \({\Rightarrow \theta=8.7^{\circ} {C}}\)
PHXI11:THERMAL PROPERTIES OF MATTER
366370
\(50\;g\) of ice at \( - 5^\circ C\) is added to \(200\;g\) of water at \(40^\circ C\) in a calorimeter. The water equivalent of calorimeter is \(50\;g\) of water. The temperature of the mixture is
1 \(9.58^\circ C\)
2 \(12.58^\circ C\)
3 \(19.58^\circ C\)
4 \(29.58^\circ C\)
Explanation:
As water equivalent of calorimeter is \(50\,g\) of water the total water present in the system is \(250\,g\). Heat absorbed by ice to become water at \(0^\circ C\) is \(Q{\rm{ }} = 50 \times 0.5 \times 5 + 50 \times 80 = 4125\,cal\) The decrease in temperature of water is \(4125 = 250 \times 1 \times \Delta T\) \( \Rightarrow \Delta T = 165^\circ C\) Now we have \(50\;g\) of water at \(0^\circ C\) and \(250\;g\) of water at \(23.5^\circ C\). The temperature of the mixture is \(50 \times 1(T - 0) + 250 \times 1 \times (T - 235) = 0\) \( \Rightarrow T = 19.58^\circ C\)
