The specific heat capacity of lead is 0.128 J/g°C. To calculate the energy required to raise the temperature of 3 kg of lead by 5 degrees Celsius, you would use the formula: energy = mass x specific heat capacity x temperature change. Therefore, the energy required would be 3 kg x 1000 g/kg x 0.128 J/g°C x 5°C = 1920 J.
1935 J (apex)
Aluminum needs less energy than lead to raise its temperature by one degree, as aluminum has a lower specific heat capacity compared to lead. This means that aluminum can absorb and release heat more easily than lead for the same change in temperature.
The specific heat capacity of lead is 0.128 J/g°C. To calculate the energy required, first convert the mass of lead to grams (1 kg = 1000 g). Then use the formula: Energy = mass x specific heat capacity x change in temperature. Plugging in the values: Energy = 3000 g x 0.128 J/g°C x (20°C - 15°C) = 1920 Joules.
The specific heat capacity of lead is 0.128 J/g°C. To calculate the heat energy needed to raise the temperature of the lead by 250°C, you would use the formula: Q = m x c x ΔT, where Q is the heat energy, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. Plugging in the values, the heat energy needed would be 6400 Joules.
1935
The specific heat capacity of lead is 0.128 J/g°C. To raise the temperature of 40 g of lead by 1°C, it requires 40 g * 0.128 J/g°C = 5.12 J of energy.
The specific heat capacity of lead is 0.128 J/g°C. To calculate the energy required to raise the temperature of 3 kg of lead by 5 degrees Celsius, you would use the formula: energy = mass x specific heat capacity x temperature change. Therefore, the energy required would be 3 kg x 1000 g/kg x 0.128 J/g°C x 5°C = 1920 J.
1935 J (apex)
Aluminum needs less energy than lead to raise its temperature by one degree, as aluminum has a lower specific heat capacity compared to lead. This means that aluminum can absorb and release heat more easily than lead for the same change in temperature.
1935 JSource: Apex
The specific heat capacity of lead is 0.128 J/g°C. To calculate the energy required, first convert the mass of lead to grams (1 kg = 1000 g). Then use the formula: Energy = mass x specific heat capacity x change in temperature. Plugging in the values: Energy = 3000 g x 0.128 J/g°C x (20°C - 15°C) = 1920 Joules.
The specific heat capacity of lead is 0.128 J/g°C. To calculate the heat energy needed to raise the temperature of the lead by 250°C, you would use the formula: Q = m x c x ΔT, where Q is the heat energy, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. Plugging in the values, the heat energy needed would be 6400 Joules.
The specific heat capacity of lead is 0.128 J/g°C. To calculate the heat energy required to melt the lead, you would first need to raise the temperature of the lead from 24°C to its melting point of 327.5°C using the equation Q = mcΔT. Then, once the lead is at its melting point, you would calculate the heat energy required to melt the lead using the equation Q = mL, where L is the heat of fusion for lead which is 23.5 kJ/kg.
A. Water would take the longest to raise its temperature compared to basalt, iron, and lead due to its high specific heat capacity, which means it requires more heat energy to raise its temperature. Basalt, iron, and lead have lower specific heat capacities and would heat up faster.
1kg of lead at 400°C would have more energy because thermal energy is directly proportional to temperature, and the higher the temperature, the higher the thermal energy.
The rate constant in the Arrhenius equation is impacted by temperature and activation energy. Increasing temperature generally increases the rate constant as molecules have more energy to overcome activation barriers. Similarly, lowering the activation energy required can lead to a higher rate constant.