24.62 g of nickel heated and then immersed in 200ml of water. the temperature of the water rose 22.50c to 32.20c.The specific heat of NI is .470J/g*C. What is the temperature of the Ni before it is immersed in the water?
[mass metal x specific heat metal x (Tf-Ti)] + [mass H2O x specific heat H2O x (Tf-Ti)] = 0
Tf is T final. Ti is T initial.
Solve for Ti metal
Thanks
To find the initial temperature of the nickel before it was immersed in water, we can use the concept of heat transfer. The heat lost by the nickel is equal to the heat gained by the water:
Heat lost by nickel = Heat gained by water
The formula for heat transfer is:
q = m * c * ΔT
Where:
q = heat transferred (in joules, J)
m = mass of the substance (in grams, g)
c = specific heat capacity of the substance (in J/g°C)
ΔT = change in temperature (in °C)
Now let's calculate the heat lost by the nickel:
q_nickel = m_nickel * c_nickel * ΔT_nickel
q_nickel = 24.62 g * 0.470 J/g°C * (T_final - T_initial)
Next, we calculate the heat gained by the water:
q_water = m_water * c_water * ΔT_water
q_water = 200 g * 1.00 J/g°C * (32.20°C - 22.50°C)
Since the heat lost by the nickel and the heat gained by the water are equal, we can set up an equation:
q_nickel = q_water
24.62 g * 0.470 J/g°C * (T_final - T_initial) = 200 g * 1.00 J/g°C * (32.20°C - 22.50°C)
Now we can plug in the given values and solve for T_initial:
24.62 g * 0.470 J/g°C * (32.20°C - T_initial) = 200 g * 1.00 J/g°C * (32.20°C - 22.50°C)
11.558 * (32.20 - T_initial) = 200 * (9.70)
11.558 * (32.20 - T_initial) = 1940
368.647 - 11.558T_initial = 1940
-11.558T_initial = 1940 - 368.647
-11.558T_initial = 1571.353
T_initial = 1571.353 / -11.558
T_initial ≈ -135.792°C
Therefore, the initial temperature of the nickel before it was immersed in the water is approximately -135.792°C.