How many degrees of temperature rise will occur when a 25.0 g block of aluminum absorbs 10.0 kJ of heat? The specific heat of Al is 0.900 J/g·°C.
To calculate the temperature rise, we can use the formula:
q = mcΔT
where:
q = heat absorbed (10.0 kJ = 10,000 J)
m = mass of aluminum (25.0 g)
c = specific heat of aluminum (0.900 J/g·°C)
ΔT = temperature rise
Plugging in the values:
10,000 J = (25.0 g)(0.900 J/g·°C)ΔT
Now, solve for ΔT:
ΔT = 10,000 J / (25.0 g * 0.900 J/g·°C)
ΔT = 10,000 J / 22.5 J/°C
ΔT = 444.44 °C
Therefore, the temperature rise will be 444.44 °C when a 25.0 g block of aluminum absorbs 10.0 kJ of heat.