Question #1dda3

Question

Question #1dda3

1 Answer

Impact of this question

6848 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-02-12T09:41:45

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, ( $T_{f} - T_{i}$ $"T"_f-"T"_i$ ).

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, $T_{f}$ $"T"_f$ , complete the following steps.

Rearrange the heat capacity equation to isolate T.

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

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 $22^{o} C$ $"22"^("o")"C"$ . Water's specific heat, c, is ${1.0 cal/g\cdot}^{o} C$ $"1.0 cal/g·"^("o")"C"$ .

$T= \frac{Q}{mc}$ $"T="Q/"mc"$ = $\frac{300.0 cal}{{100.0g\cdot1.0 cal/g\cdot}^{o} C}$ $"300.0 cal"/("100.0g·1.0 cal/g·"^("o")"C"$ = ${3.0}^{o} C$ $"3.0"^("o")"C"$

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

$T_{f}$ $"T"_f"$ = $T_{i}$ $"T"_i"$ + $T$ $"T"$ = $22^{o} C$ $"22"^("o")"C"$ + ${3.0}^{o} {C = 25}^{o} C$ $"3.0"^("o")"C = 25"^("o")"C"$

Science

Math

Humanities

... and beyond

Question #1dda3

1 Answer

Related questions

Impact of this question