# We add 100 mL of water at 20 degrees C to 10 g of an unknown substance which is at 50 degrees C....

## Question:

We add 100 mL of water at 20 degrees C to 10 g of an unknown substance which is at 50 degrees C. The water ends up at 25 degrees C. How much heat was transferred?

## Heat Transfer:

The process of heat transfer results to the change in the temperature of a substance. We can quantify the absorbed heat using the equation, {eq}\displaystyle q = mc\Delta T {/eq}, where m is the mass of the substance, c is its specific heat, and {eq}\displaystyle \Delta {/eq}T is the change in temperature. This equation holds true when no phase change nor chemical change occurs.

Determine the total heat transferred, q, to the water by applying the equation, {eq}\displaystyle q = mc(T_f-T_i) {/eq}, where m is the mass of the water, c = 1 cal/g{eq}\displaystyle ^\circ {/eq}C is the specific heat, {eq}\displaystyle T_i {/eq} = 20{eq}\displaystyle ^\circ {/eq}C, and {eq}\displaystyle T_f {/eq} = 25{eq}\displaystyle ^\circ {/eq}C. We acquire the mass of water by multiplying the volume of water, V = 100 mL, to the known density, {eq}\displaystyle \rho {/eq} = 1 g/mL, or {eq}\displaystyle m = V\times \rho {/eq}. We proceed with the solution.

{eq}\begin{align} \displaystyle q &= mc(T_f-T_i)\\ &= V\times \rho\times c(T_f-T_i)\\ &= 100\ mL\times 1\ g/mL\times 1\ cal/g^\circ C \times (25 ^\circ C- 20 ^\circ C)\\ &= 500\ cal \end{align} {/eq} 