To calculate the heat given off, we need to use the formula:
q = m × c × ΔT
where:
q = heat given off
m = mass of copper (104.1 g)
c = specific heat capacity of copper (0.385 J/g °C)
ΔT = change in temperature (final temperature - initial temperature)
By substituting the given values:
ΔT = 23.0 °C - 155.0 °C = -132.0 °C (we get a negative value because the temperature is decreasing)
q = 104.1 g × 0.385 J/g°C × -132.0 °C
q = -5181.78 J
Therefore, the heat given off is -5181.78 J (or you can simply write it as 5182 J, since the negative sign indicates that heat is being released).
Calculate the heat given off when 104.1 g of copper cools from 155.0oC to 23.0oC. The specific heat capacity of copper is 0.385 J/g oC.
5 answers
a
The heat given off when 104.1 g of copper cools from 155.0°C to 23.0°C is 5182 J.
how do you know this
To calculate the heat given off, we can use the formula:
q = m × c × ΔT
where:
q = heat given off
m = mass of copper (104.1 g)
c = specific heat capacity of copper (0.385 J/g °C)
ΔT = change in temperature (final temperature - initial temperature)
By substituting the given values:
ΔT = 23.0 °C - 155.0 °C = -132.0 °C (we get a negative value because the temperature is decreasing)
q = 104.1 g × 0.385 J/g °C × -132.0 °C
q = -5181.78 J
Therefore, the heat given off is -5181.78 J (or we can write it as 5182 J, since the negative sign indicates that heat is being released).
q = m × c × ΔT
where:
q = heat given off
m = mass of copper (104.1 g)
c = specific heat capacity of copper (0.385 J/g °C)
ΔT = change in temperature (final temperature - initial temperature)
By substituting the given values:
ΔT = 23.0 °C - 155.0 °C = -132.0 °C (we get a negative value because the temperature is decreasing)
q = 104.1 g × 0.385 J/g °C × -132.0 °C
q = -5181.78 J
Therefore, the heat given off is -5181.78 J (or we can write it as 5182 J, since the negative sign indicates that heat is being released).
3 answers