366354 A piece of ice falls from a height \(h\) so that it melts completely. Only half of the heat produced is absorbed by the ice. The value of \(h\) is (Latent heat of ice is \(3.4 \times {10^5}\;J/kg\) and \(g = 10\;N/kg)\)

366355 Oil at \(25^\circ C\) is poured very slowly into a calorimeter that is at temperature of \(120^\circ C.\) The boiling point of the oil is \(80^\circ C.\) It is found that the first \(5\;g\) of the oil completely evaporates. After pouring another \(75\;g\) of the oil the equilibrium temperature is found to be \(40^\circ C.\) The ratio of the latent heat of the oil to its speific heat will be
(Neglect the heat exchange with surrounding)

366356 It is difficult to cook rice in an open vessel by boiling it a high altitudes because of

366357 Sunrays are allowed to fall on a long of diameter \(20\,cm\). They are then brought to focus on a calorimeter containing \(20\,g\) of ice. If the absorption by the lens is negligible, the time required to melt all the ice is (solar constant \(=\) \(1.9/cal/\min /c{m^2}\) and \(L = 80\,cal/g\))

366358 \('M'{\rm{ }}kg\) of water at \(80^\circ C\) is divided into two parts so that one part of mass \('m'{\rm{ }}kg\) when converted into ice at \(0^\circ C\) would release enough heat to vaporise the other part entirely, then \(\frac{m}{M}\) is equal to
(Take Specific heat of water \( = 1{\mkern 1mu} \,{\rm{cal}}\,{g^{ - 1}}\,^\circ {C^{ - 1}},\) Latent heat of fusion of ice \( = 80\,{\rm{cal}}\,\,{g^{ - 1}},\) Latent heat of steam \( = 540\,{\rm{cal}}\,\,{g^{ - 1}}\))

366354 A piece of ice falls from a height \(h\) so that it melts completely. Only half of the heat produced is absorbed by the ice. The value of \(h\) is (Latent heat of ice is \(3.4 \times {10^5}\;J/kg\) and \(g = 10\;N/kg)\)

366355 Oil at \(25^\circ C\) is poured very slowly into a calorimeter that is at temperature of \(120^\circ C.\) The boiling point of the oil is \(80^\circ C.\) It is found that the first \(5\;g\) of the oil completely evaporates. After pouring another \(75\;g\) of the oil the equilibrium temperature is found to be \(40^\circ C.\) The ratio of the latent heat of the oil to its speific heat will be
(Neglect the heat exchange with surrounding)

366356 It is difficult to cook rice in an open vessel by boiling it a high altitudes because of

366357 Sunrays are allowed to fall on a long of diameter \(20\,cm\). They are then brought to focus on a calorimeter containing \(20\,g\) of ice. If the absorption by the lens is negligible, the time required to melt all the ice is (solar constant \(=\) \(1.9/cal/\min /c{m^2}\) and \(L = 80\,cal/g\))

366358 \('M'{\rm{ }}kg\) of water at \(80^\circ C\) is divided into two parts so that one part of mass \('m'{\rm{ }}kg\) when converted into ice at \(0^\circ C\) would release enough heat to vaporise the other part entirely, then \(\frac{m}{M}\) is equal to
(Take Specific heat of water \( = 1{\mkern 1mu} \,{\rm{cal}}\,{g^{ - 1}}\,^\circ {C^{ - 1}},\) Latent heat of fusion of ice \( = 80\,{\rm{cal}}\,\,{g^{ - 1}},\) Latent heat of steam \( = 540\,{\rm{cal}}\,\,{g^{ - 1}}\))

366354 A piece of ice falls from a height \(h\) so that it melts completely. Only half of the heat produced is absorbed by the ice. The value of \(h\) is (Latent heat of ice is \(3.4 \times {10^5}\;J/kg\) and \(g = 10\;N/kg)\)

366355 Oil at \(25^\circ C\) is poured very slowly into a calorimeter that is at temperature of \(120^\circ C.\) The boiling point of the oil is \(80^\circ C.\) It is found that the first \(5\;g\) of the oil completely evaporates. After pouring another \(75\;g\) of the oil the equilibrium temperature is found to be \(40^\circ C.\) The ratio of the latent heat of the oil to its speific heat will be
(Neglect the heat exchange with surrounding)

366356 It is difficult to cook rice in an open vessel by boiling it a high altitudes because of

366357 Sunrays are allowed to fall on a long of diameter \(20\,cm\). They are then brought to focus on a calorimeter containing \(20\,g\) of ice. If the absorption by the lens is negligible, the time required to melt all the ice is (solar constant \(=\) \(1.9/cal/\min /c{m^2}\) and \(L = 80\,cal/g\))

366358 \('M'{\rm{ }}kg\) of water at \(80^\circ C\) is divided into two parts so that one part of mass \('m'{\rm{ }}kg\) when converted into ice at \(0^\circ C\) would release enough heat to vaporise the other part entirely, then \(\frac{m}{M}\) is equal to
(Take Specific heat of water \( = 1{\mkern 1mu} \,{\rm{cal}}\,{g^{ - 1}}\,^\circ {C^{ - 1}},\) Latent heat of fusion of ice \( = 80\,{\rm{cal}}\,\,{g^{ - 1}},\) Latent heat of steam \( = 540\,{\rm{cal}}\,\,{g^{ - 1}}\))

366354 A piece of ice falls from a height \(h\) so that it melts completely. Only half of the heat produced is absorbed by the ice. The value of \(h\) is (Latent heat of ice is \(3.4 \times {10^5}\;J/kg\) and \(g = 10\;N/kg)\)

366355 Oil at \(25^\circ C\) is poured very slowly into a calorimeter that is at temperature of \(120^\circ C.\) The boiling point of the oil is \(80^\circ C.\) It is found that the first \(5\;g\) of the oil completely evaporates. After pouring another \(75\;g\) of the oil the equilibrium temperature is found to be \(40^\circ C.\) The ratio of the latent heat of the oil to its speific heat will be
(Neglect the heat exchange with surrounding)

366356 It is difficult to cook rice in an open vessel by boiling it a high altitudes because of

366357 Sunrays are allowed to fall on a long of diameter \(20\,cm\). They are then brought to focus on a calorimeter containing \(20\,g\) of ice. If the absorption by the lens is negligible, the time required to melt all the ice is (solar constant \(=\) \(1.9/cal/\min /c{m^2}\) and \(L = 80\,cal/g\))

366358 \('M'{\rm{ }}kg\) of water at \(80^\circ C\) is divided into two parts so that one part of mass \('m'{\rm{ }}kg\) when converted into ice at \(0^\circ C\) would release enough heat to vaporise the other part entirely, then \(\frac{m}{M}\) is equal to
(Take Specific heat of water \( = 1{\mkern 1mu} \,{\rm{cal}}\,{g^{ - 1}}\,^\circ {C^{ - 1}},\) Latent heat of fusion of ice \( = 80\,{\rm{cal}}\,\,{g^{ - 1}},\) Latent heat of steam \( = 540\,{\rm{cal}}\,\,{g^{ - 1}}\))

366354 A piece of ice falls from a height \(h\) so that it melts completely. Only half of the heat produced is absorbed by the ice. The value of \(h\) is (Latent heat of ice is \(3.4 \times {10^5}\;J/kg\) and \(g = 10\;N/kg)\)

366355 Oil at \(25^\circ C\) is poured very slowly into a calorimeter that is at temperature of \(120^\circ C.\) The boiling point of the oil is \(80^\circ C.\) It is found that the first \(5\;g\) of the oil completely evaporates. After pouring another \(75\;g\) of the oil the equilibrium temperature is found to be \(40^\circ C.\) The ratio of the latent heat of the oil to its speific heat will be
(Neglect the heat exchange with surrounding)

366356 It is difficult to cook rice in an open vessel by boiling it a high altitudes because of

366357 Sunrays are allowed to fall on a long of diameter \(20\,cm\). They are then brought to focus on a calorimeter containing \(20\,g\) of ice. If the absorption by the lens is negligible, the time required to melt all the ice is (solar constant \(=\) \(1.9/cal/\min /c{m^2}\) and \(L = 80\,cal/g\))

366358 \('M'{\rm{ }}kg\) of water at \(80^\circ C\) is divided into two parts so that one part of mass \('m'{\rm{ }}kg\) when converted into ice at \(0^\circ C\) would release enough heat to vaporise the other part entirely, then \(\frac{m}{M}\) is equal to
(Take Specific heat of water \( = 1{\mkern 1mu} \,{\rm{cal}}\,{g^{ - 1}}\,^\circ {C^{ - 1}},\) Latent heat of fusion of ice \( = 80\,{\rm{cal}}\,\,{g^{ - 1}},\) Latent heat of steam \( = 540\,{\rm{cal}}\,\,{g^{ - 1}}\))