No. 1 2 Substance Water R134a T °C 300 P kPa 1300 m³/kg u kJ/kg h kJ/kg S kJ/kg-K 0.37973 Quality X 0

SolVe correctly please. There are 500 liquid medicines, 100 grams each, in a box. In how many minutes can a refrigerator with a power of 0.45 kW and an efficiency coefficient of 2.5 cool the medicines from 20 ° C to 8 ° C? The specific heat of the drug is given as 4.2 kJ/kg °C.

If the mass of the string and the cup are so small that they are negligible, what is the approximate minimum number of 10-cent coins needed to break a bar of chocolate that has a flexural stress of 515,000 Pa? The length, thickness, and width measurements of the chocolate bar are 12mm each, and the mass of one 10-cent coin is 2.5g. Answer : zero, one, two or five?

MATCH THE ANSWER WITH THE QUESTION Path Function Choose. Choose. Isobaric Process Pressure is constant Нeat Intensive property No work done Pressure Adiabatic Process Volume No transfer of heat

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Completely solve and box the final answer. Write legibly 4. Compute the heat conducted in KW through a sheet of plate glass, k = 0.025 Cal/s-cm-ºC, which is 3.5 m x 4.8 m and 12.5mm thick. The temperature of the surfaces are 28ºC and negative 5.4ºC.

The pressure increase of a gas was divided into three calculation paths; Arrange the temperature of each path in increasing magnitude: Isothermal compression(T1) isochoric heating(T2 adiabatic compression(T3). a.T2<T1<T3 b.T1<T2<T3 c.T1<T3<T2 d.T2<T3<T1

The specific heat of copper is 0.093 cal/(g.°C) and the specific heat of gold is 0.031 cal/(g.°C). If 5.8 cal is supplied to one gram of copper and one gram of gold, the RATIO of temperature increase of gold to that of copper is

Solve it correctly please. I will rate accordingly with 4votes.

number 1 A food product containing 82% moisture content is being frozen. Estimate the specific heat of the product at -8 ° C when 82% of the water is frozen. The specific heat of the dry product is 2.5 kJ / (kg ° C). It is assumed that the specific heat of water at -10 ° C is the same as the specific heat of water at 0 ° C, and that the specific heat of ice follows the function Cp es = 0.0062 T Frozen + 2.0649. Cp of frozen product = kJ / kg ° C.