Answer:
207.50°C
Explanation:
Mass of metal (M1) = 440g
Volume of water = 400mL
Initial temperature of water (T2) = 20°C
Final temperature of the mixture (T3) = 35.80°C
Specific heat capacity of the metal (C1) = 0.3850J/g°C
Specific heat capacity of water = 4.184J/g°C
Density of water = 1g/mL
Temperature of metal (T1) = ?
Heat lost by the metal = Heat gained by the water
Q = MC∇T
Q = heat energy
M = Mass
C = specific heat capacity
∇T = change in temperature
M1C1(T1 - T3) = M2C2(T3 - T2)
M2 = ?
Density = mass / volume
Mass = density * volume
Mass = 1g/mL * 400mL
Mass = 400g
M1C1(T1 - T3) = M2C2(T3 - T2)
400 * 0.3850 (T1 - 35.80) = 400 * 4.184 * (35.80 - 20)
154(T1 - 35.80) = 1673.6 * (15.8)
154T1 - 5513.2 = 26442.88
154T1 = 26442.88 + 5513.2
154T1 = 31956.08
T1 = 31956.08 / 154
T1 = 207.50
The initial temperature of the metal was 207.50°C
100 degrees C
c.
-183 degrees C
b.
-253 degrees C
d.
0 degrees C
Answer:
Explanation:
Answer:
The resistance would be 64.10 ohms.
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