Answer:
Explanation:
There are two heat transfers to consider:
Calculate q₂
m = 120.0 g
C = 4.184 J·°C⁻¹g⁻¹
T₂ = 29.20 °C
T₁ = 20.00 °C
ΔT = T₂ - T₁ =(29.20 – 20.00) °C =9.20 °C
q₂ = 120.0 g × 4.184 J·°C⁻¹g⁻¹ × 9.20 °C = 4620 J = 4.62 kJ
Calculate q₁
q₁ + 4.62 kJ = 0
q₁ = -4.62 kJ
The negative sign shows that heat is given off.
Answer:
The mass of a sample of iron that has had 300 J applied to it and heats up from 20 degrees Celsius to 40 degrees Celsius is 32.61 grams.
Explanation:
Calorimetry is the measurement and calculation of the measurement of heat changes exchanged by a body or a system produced in physical and chemical processes.
The sensible heat of a body is the amount of heat received or transferred by a body to produce a change in temperature but without a change in physical state.
The sensible heat in a constant pressure is calculated by:
Q = c * m * ΔT
where Q is the heat exchanged by a body of mass m, constituted by a substance of specific heat c, and where ΔT is the temperature variation (ΔT=Tfinal - Tinitial)
In this case:
Replacing:
300 J= 0.46 * m* (40 - 20) C
Solving:
300 J= 0.46 * m* 20 C
m= 32.61 g