# Question #1dda3

Feb 12, 2015

I believe this is a heat capacity problem.

The equation for heat capacity is Q=cmT, where Q = the heat lost or gained, c = specific heat, m = mass in grams, and T = the temperature change, (${\text{T"_f-"T}}_{i}$).

The density and volume are used to calculate the mass, most likely of water. Mass = density x volume.

In order to determine the final temperature, ${\text{T}}_{f}$, complete the following steps.

Rearrange the heat capacity equation to isolate T.

$\text{T = "Q/"mc}$

Now plug in your known values to get T, the change in temperature.

For example, suppose 300.0 calories were added to 100.0 grams of water at an initial temperature of $\text{22"^("o")"C}$. Water's specific heat, c, is $\text{1.0 cal/g·"^("o")"C}$.

$\text{T="Q/"mc}$ = $\text{300.0 cal"/("100.0g·1.0 cal/g·"^("o")"C}$ = $\text{3.0"^("o")"C}$

The initial temperature, $\text{T"_i}$ was $\text{22"^("o")"C}$. Now add the change in temperature, T = $\text{3.0"^("o")"C}$ to the initial temperature and that is the final temperature.

$\text{T"_f}$ = $\text{T"_i}$ + $\text{T}$ = $\text{22"^("o")"C}$ + $\text{3.0"^("o")"C = 25"^("o")"C}$