A 33.0g sample of glass, which has a specific heat capacity of 0.670·J·g−1°C−1, is dropped into an insulated container containing 200.0g of water at 70.0°C and a constant pressure of 1atm. The initial temperature of the glass is 7.6°C. Assuming no heat is absorbed from or by the container, or the surroundings, calculate the equilibrium temperature of the water. Be sure your answer has 3 significant digits.

68.4°C is the final temperature of the water

Explanation:

In this problem, the heat lost by the water is equal to the heat lost by the sample of glass, that is:

W(Gl)*S(Gl)*ΔT(Gl) = W(H₂O)*S(H₂O)*ΔT(H₂O)

Where W is mass (Gl = 33.0g; H₂O = 200.0g)

S is specific heat (Gl = 0.670J/g°C; H₂O = 4.184J/g°C)

And ΔT is change in temperature ( Gl = T-7.6°C; H₂O = 70.0°C-T)

Replacing:

33.0g*0.670J/g°C*(T-7.6°C) = 200.0g*4.184J/g°C*(70.0°C-T)

22.11*(T-7.6) = 836.8*(70.0-T)

22.11T - 168.04 = 58576 - 836.8T

858.91T = 58744.04

T = 68.4°C is the final temperature of the water

girl imma just put my comment here because i need a couple point okurrttt