If you have 330.0 mL of water at 25.00 °C and add 100.0 mL of water at 95.00 °C, what is the final temperature of the mixture? Use 1.00 g/mL as the density of water.

Question

aaronq · Answer

the thermal energy (heat) lost but the hot water is transferred to the cold water, (assuming that loss does not occur to any other object) we can say: $-q_{hot}=q_{warm}$
from the calorimetry equation "$q=m*c_p*\Delta T$",

we can get:   $-[m_1\Delta T]=m_2\Delta T\rightarrow -[m_1(T_f-T_i)]=m_2\Delta(T_f-T_i)$

because the heat capacities are equal (they are both liquid water), we can ignore it.

aaronq · Answer

the last part shouldn't have a $\Delta$ sign.  
it should read: "$-[m_1(T_f-T_i)]=m_2(T_f-T_i) $"

anonymous · Answer

so wait, does that mean that finding water's specific heat is not necessary?

aaronq · Answer

no because they're both liquid water, so the $C_p$ will just cancel out in the math.

anonymous · Answer

so by inserting the data, the equation would be -(330.0(T-25))=100(t-95)?

aaronq · Answer

yep

anonymous · Answer

and then just solve for T to find the final temperature?

aaronq · Answer

yep