Answer to Question #265150 in Chemistry for Jose

Question

When 1.00 kg lead specific heat =0.13jg °c at 100.0°c is adde to a quantity of water at 28.5, The final temperature of the lead water mixture is 35.2°c what is the mass of water present?

Answer 1

Solution:

This problem can be summarized thusly:

q_{lost by lead} = -q_{gained by water}

 

q = m × C × ΔT

q = m × C × (T_f - T_i),

where:

q = amount of heat energy gained or lost by substance (J)

m = mass of sample (g)

C = specific heat capacity (J °C^-1 g^-1)

T_f = final temperature (°C)

T_i = initial temperature (°C)

C_water = 4.186 J °C^-1 g^-1

C_lead = 0.130 J °C^-1 g^-1

Therefore,

q_{lost by lead} = (1000 g) × (0.130 J °C^-1 g^-1) × (35.2 - 100.0)°C = -8424 J

q_{lost by lead} = -8424 J

q_{gained by water} = m_water × (4.186 J °C^-1 g^-1) × (35.2 - 28.5)°C = m_water × 28.0462 J

q_{gained by water} = m_water × 28.0462 J

q_{lost by lead} = -q_{gained by water}

-8424 = -28.0462 × m_water

m_water = (8424) / (28.0462) = 300.36 g = 300 g

m_water = 300 g

Answer: The mass of water is 300 grams