How much energy must be transferred to raise the temperature of a cup of coffee?
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How much energy must be transferred to raise the temperature of a cup of coffee?
The energy required is 77.5 kJ.
How do you calculate the energy needed to cool water?
The specific heat capacity is the heat or energy required to change one unit mass of a substance of a constant volume by 1 °C. The formula is Cv = Q / (ΔT ⨉ m) .
How much energy does it take to make a cup of coffee?
Depending on the model, it takes about 200 to 400 watts for every cup, all things considered. Add the idle power consumption, and it’s safe to estimate with an average of 300 kilowatt per cup. Now, compare that with an automated drip coffee maker.
What is the energy required to heat water?
The specific heat capacity of water is 4,200 Joules per kilogram per degree Celsius (J/kg°C). This means that it takes 4,200 J to raise the temperature of 1 kg of water by 1°C.
How much energy does it take to heat a cup of coffee?
How much energy must be transferred to raise the temperature of a cup of coffee (250 mL) from 20.5 C (293.7 K) to 95.6 C (368.8 K)? Assume that water and coffee have the same density (1.00 g/ mL), and specific heat capacity 4.184 J/ gK)? The energy required is 77.5 kJ.
Is a cup of coffee at 100 degrees too hot to drink?
You have a 200 gram cup of coffee at 100 °C, too hot to drink. Various cooling strategies demonstrate specific heat, phase changes, and the approach to thermal equilibrium. How much will the coffee be cooled by adding 50 gm of water at 0 °C?
What is the specific heat capacity value of copper?
What is the specific heat capacity value of copper? The specific heat of copper is 385 J/kg K. You can use this value to estimate the energy required to heat a 100 g of copper by 5 °C, i.e., Q = m x Cp x ΔT = 0.1 * 385 * 5 = 192.5 J. What is the specific heat capacity value of aluminum?
Is it possible to extract 25000 calories from coffee at 0°C?
But this can’t be right because it gives a negative temperature (-8°C) and the specific heat equation Q = cmΔT is valid only so long as a phase change is not encountered, so we can’t pass 0°C with this equation. If 25000 calories are extracted, we have cooled the coffee to 0°C but still have 2000 cal to remove. This will freeze some of the coffee: