Answer:
68.4°C is the final temperature of the water
Step-by-step 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