Question

What is the specific heat capacity of silver metal if 60.00 g of the metal absorbs 40.3J of heat and the temperature rises 25°C.

Q = m-c-AT, Ç=Q/m.AT)
O 0.02686 J/g °C
O 0.02686 cal/g °C
O 0.027 J/g °C
O 0.0268 J/g °C

Answer 1

The specific heat capacity of silver metal, when 60.00 g absorbs 40.3 J of heat and the temperature increase is 25°C, is calculated to be 0.02686 J/g°C.

Step-by-step explanation:

The specific heat capacity of silver metal can be determined using the provided information and the formula № = Q / (m ⋅ ΔT). In this formula, № is the specific heat capacity, Q is the quantity of heat absorbed, m is the mass of the substance, and ΔT is the change in temperature. Given that 60.00 g of silver absorbs 40.3 J of heat and the temperature rises by 25°C, the specific heat capacity № can be calculated as:

№ = Q / (m ⋅ ΔT) = 40.3 J / (60.00 g ⋅ 25°C) = 40.3 J / (1500 g°C) = 0.02686 J/g°C.