If a 50.0g silver spoon at 20.0 degrees C. How much heat does the spoo absorb from the coffee to reach a temperature of 89.0degrees C?
hey im thinking the answer to this can be found in the first law of therodynamics
JQ-w=u2 - u1
To calculate the heat absorbed by the silver spoon, you can use the formula:
Q = mcΔT
Where:
Q is the heat absorbed (in joules)
m is the mass of the spoon (in grams)
c is the specific heat capacity of silver (in joules per gram per degree Celsius)
ΔT is the change in temperature (in degrees Celsius)
First, let's calculate the change in temperature of the spoon:
ΔT = final temperature - initial temperature
= 89.0°C - 20.0°C
= 69.0°C
Next, let's find the specific heat capacity of silver. The specific heat capacity of silver is approximately 0.235 J/g°C.
Now, we can substitute the values into the formula:
Q = (50.0g) * (0.235 J/g°C) * (69.0°C)
= 801.75 J
Therefore, the silver spoon absorbs approximately 801.75 joules of heat from the coffee to reach a temperature of 89.0°C.