Answer:
D = 28.2g
Explanation:
Initial temperature of metal (T1) = 155°C
Initial Temperature of calorimeter (T2) = 18.7°C
Final temperature of solution (T3) = 26.4°C
Specific heat capacity of water (C2) = 4.184J/g°C
Specific heat capacity of metal (C1) = 0.444J/g°C
Volume of water = 50.0mL
Assuming no heat loss
Heat energy lost by metal = heat energy gain by water + calorimeter
Heat energy (Q) = MC∇T
M = mass
C = specific heat capacity
∇T = change in temperature
Mass of metal = M1
Mass of water = M2
Density = mass / volume
Mass = density * volume
Density of water = 1g/mL
Mass(M2) = 1 * 50
Mass = 50g
Heat loss by the metal = heat gain by water + calorimeter
M1C1(T1 - T3) = M2C2(T3 - T2)
M1 * 0.444 * (155 - 26.4) = 50 * 4.184 * (26.4 - 18.7)
0.444M1 * 128.6 = 209.2 * 7.7
57.0984M1 = 1610.84
M1 = 1610.84 / 57.0984
M1 = 28.21g
The mass of the metal is 28.21g
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