q=mLvamericanforest said:What formula do you know that is related to heating and/or transforming material from one physical state to another? Those are the two processes which must occur here.
Yes.MaryBarnes said:i attempted the answer:
Q total=mcwΔt+msLf
=8.0kg*4200j/kg*(100°C-25°C)+2.0kg*2.3*106
=7.12*106J
awesome! thanks for the help!haruspex said:Yes.
If you want to make your working letter-perfect, include all units: 8.0kg*4200J/kg/°C*(100°C-25°C)+2.0kg*2.3*106J/kg.
The density of water at 25degC is 997 kg/m^3. Therefore, the density of 8.0kg of water in the container is 997 kg/m^3.
The volume of water in the container can be calculated using the formula V = m/d, where V is the volume, m is the mass, and d is the density. In this case, the volume of water is 0.0080 m^3.
The specific heat capacity of water is 4.186 J/g*degC. Therefore, the total heat energy required to raise the temperature of 8.0kg of water from 25degC to 30degC is 167.44 kJ.
The pressure inside the container can be calculated using the ideal gas law, PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature. In this case, the pressure inside the container is equal to the atmospheric pressure, which is approximately 101.3 kPa.
The time it takes for the water to reach 30degC depends on various factors such as the heating method, the material of the container, and the surrounding temperature. However, assuming the water is heated with a constant power source, the time can be calculated using the formula t = mcΔT/Q, where t is the time, m is the mass, c is the specific heat capacity, ΔT is the change in temperature, and Q is the power. In this case, the time it takes for the water to reach 30degC will be approximately 8.2 minutes.