1125 J of energy is used to heat 250 g of iron to 55 °C. The specific heat capacity of iron is 0.45 J/(g·°C).
What was the temperature of the iron before it was heated?
To find the initial temperature of the iron before it was heated, we can use the equation:
Q = m * c * ΔT
where:
Q is the energy transferred (in Joules),
m is the mass of the substance (in grams),
c is the specific heat capacity of the substance (in J/(g·°C)), and
ΔT is the change in temperature (final temperature - initial temperature) (in °C).
In this case, we are given:
Q = 1125 J
m = 250 g
c = 0.45 J/(g·°C)
ΔT = 55 °C
Rearranging the equation to solve for initial temperature (T_initial):
Q = m * c * ΔT
1125 J = 250 g * 0.45 J/(g·°C) * 55 °C
1125 J = 250 g * 0.45 J/g * 55 °C
1125 J = 250 g * 24.75 J
1125 J = 6187.5 J
Now, we can solve for T_initial:
T_initial = ΔT + T_final
T_initial = -55 °C + T_final
Substituting the values we know:
T_initial = -55 °C + (55 °C) (since we are given that the final temperature is 55 °C)
T_initial = 0 °C
Therefore, the initial temperature of the iron before it was heated was 0 °C.
1125 = 250 * 0.45 * (55 - t)
solve for t