A solar collector receiving solar radiation at the rate of `0.6``kW`/`m^(2)`transforms it to the internal energy of a fluid at an overall efficiency of 50%. The fluid heated to 350K is used to run a heat engine which rejects heat at 313K .If the heat engine is to deliver 2.5kW power,the minimum area of the solar collector required would be

Imagine a system, that can keep the room temperature within a narrow range between 20^@C to 25^@C . The system includes a heat engine operating with variable power P = 3KT, where K is a constant coefficient, depending upon the thermal insulation of the room, the area of the walls and the thickness of the walls. T is temperature of the room in temperature drops lower than 20^@C , the engine turns on, when the temperature increase over 25^@ C, the engine turns off, room looses energy at a rate of K(T - T_0), T_0 is the outdoor temperature. The heat capacity of the room is C. Given (T_0=10^@C, ln(3/2) =0.4 , ln(6/5)=0.18 , C/K =750 SI-unit) Suppose at t = 0, the engine turns off, after how much time interval, again, the engine will turn on

Imagine a system, that can keep the room temperature within a narrow range between 20^@C to 25^@C . The system includes a heat engine operating with variable power P = 3KT, where K is a constant coefficient, depending upon the thermal insulation of the room, the area of the walls and the thickness of the walls. T is temperature of the room in temperature drops lower than 20^@C , the engine turns on, when the temperature increase over 25^@ C, the engine turns off, room looses energy at a rate of K(T - T_0), T_0 is the outdoor temperature. The heat capacity of the room is C. Given (T_0=10^@C, ln(3/2) =0.4 , ln(6/5)=0.18 , C/K =750 SI-unit) Suppose at t = 0, the engine turns on, after how much time interval again, the engine will turn off

A large family uses 8KW of power. Direct solar energy is incident on the horizontal surface at an average rate of 200Wm^-2 . If 20% of this energy can be converted to useful electrical energy, how large an area is needed to supply 8kW ?

A water cooler of storages capacity 120 liters can cool water at a constant rate of P watts. In a closed circulation system (as shown schematically in the figure), the water from the cooler is used to cool an external device that generates constantly 3kW of heat (thermal load). The temperature of water fed into the device cannot exceed 30^@C and the entire stored 120 liters of water is initially cooled to 10^@C. The entire system is thermally insulated. The minimum value of P ( in watts) for which the device can be operated for 3hours is (Specific heat of water is 4.2kJkg^-1K^-1 and the density of water is 1000kgm^-3 )

A family enters a winter vacation cabin whose walls are adiabatic, initially the interior temperature is the same as the outside temperature (0^(@)C) . The cabin consists of a single room of floor area 6m by 4m and height 3m . The room contains a 2kW electric heater. Assuming that the room is perfectly airlight and that all the heat from the heater is absorbed by the air, none escaping through the walls. The time after the heater is turned on, the air temperature reaches the comfort level of 21^(@)C in x xx100 sec , then find x (Take C_(v)=20.8 J//mol-K )

A fuel cell is a cell that is continously supplied with an oxidant and a reductant so that if can deliver a current indefinitely. Fuel cells offer the possibility of achieving high thermodynamic efficiency in the conversion of Gibbs energy into mechanical work.Internal combustion engines at best convert only the fraction (T_2-T_1)//T_2 of the heat of combustion into mechanical work. While the thermodynamic efficiency of the fuel cell is given by, eta=(DeltaG)/(DeltaH) , where DeltaG is the Gibbs energy change for the cell reaction and DeltaH is the enthalpy change of the cell reaction.A hydrogen-oxygen fuel cell may have an acidic or alkaline electrolyte. Pt|H_2(g)|H^(+)(aq.)||H_2O(l)|O_(2)(g)|Pt , (2.303 RT)/F=0.06 The above fuel cell is used to produce constant current supply under constant temperature & 30 atm constant total pressure conditions in a cylinder.If 10 moles H_2 and 5 moles of O_2 were taken initially. Rate of consumption of O_2 is 10 milli moles per minute. The half-cell reactions are 1/2O_2(g)+2H^+(aq)+2e^(-)toH_2O(l) E^(@)=1.246 V 2H^+(aq)+2e^(-) to H_2(g) E^(@)=0 To maximize the power per unit mass of an electrochemical cell, the electronic and electrolytic resistances of the cell must be minimized.Since fused salts have lower electolytic resistances than aqueous solutions, high-temperature electrochemical cells are of special interest for practical applications. Calculate e.m.f of the given cell at t=0.(log 2=0.3)

Liquid water coats an active (growing) icicle and extends up a short, narrow tube along the central axis. Because the water-ice interface must have a temperature of 0^(@) C, the water in the tube cannot lose energy through the sides of the icicle or down through the tip because there is no temperature change in those directions . It can lose energy and freeze only by sending energy up(through distance L) to the top of the icicle, where the temperature T_(r) can be below 0^(@) C. Take L=0.10m and T_(r) = -5^(@) C.Assume that the central tube and the upward conduction path both have crosse-sectional area A=0.5 m^(2) . The thermal conductivity of ice is 0.40 W //mcdotK , latent heat of fusion is L_(F)=4.0xx10^(5) J//K and the density of liquid water is 1000 kg //m^(3) . The rate at which mass converted from liquid to ice at the top of the central tube is

Liquid water coats an active (growing) icicle and extends up a short, narrow tube along the central axis. Because the water-ice interface must have a temperature of 0^(@) C, the water in the tube cannot lose energy through the sides of the icicle or down through the tip because there is no temperature change in those directions . It can lose energy and freeze only by sending energy up(through distance L) to the top of the icicle, where the temperature T_(r) can be below 0^(@) C. Take L=0.10m and T_(r) = -5^(@) C.Assume that the central tube and the upward conduction path both have crosse-sectional area A=0.5 m^(2) . The thermal conductivity of ice is 0.40 W //mcdotK , latent heat of fusion is L_(F)=4.0xx10^(5) J//K and the density of liquid water is 1000 kg //m^(3) . At what rate does the top of the tube move downward because of water freezing there?

Earth receives 1400 W//m^2 of solar power. If all the solar energy falling on a lens of area 0.2m^2 is focused on to a block of ice of mass 280 grams, the time taken to melt the ice will be….. Minutes. (Latent heat of fusion of ice= 3.3xx10^5J//kg)

Earth receives 1400 W//m^2 of solar power. If all the solar energy falling on a lens of area 0.2m^2 is focused on to a block of ice of mass 280 grams, the time taken to melt the ice will be….. Minutes. ( Latent heat of fusion of ice = 3.3xx10^5J//kg .)