Final answer:
Approximately 1,582.9 Joules of heat is needed to raise the temperature of 150 g of manganese by 22°C, using the formula q = m × c × ΔT and the specific heat capacity for manganese (0.479 J/g°C).
Step-by-step explanation:
The question asks how much heat, in Joules, is required to raise the temperature of 150 g of manganese by 22°C. To calculate the heat (q) we use the formula q = m × c × ΔT, where m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature. While the specific heat capacity of manganese is not given in the question, it is typically about 0.479 J/g°C. Therefore, using this value, the heat required is calculated as follows:
Heat (q) = 150 g × 0.479 J/g°C × 22°C
Calculating this yields:
Heat (q) = 1,582.9 Joules
Thus, 1,582.9 Joules of heat is needed to raise the temperature of 150 g of manganese by 22°C.