Answer:
[tex]\boxed {\boxed {\sf 1.87 \J/g \textdegree C}}[/tex]
Explanation:
We are asked to find the specific heat capacity of a liquid. We are given the heat added, the mass, and the change in temperature, so we will use the following formula.
[tex]q= mc\Delta T[/tex]
The heat added (q) is 47.1 Joules. The mass (m) of the liquid is 14.0 grams. The specific heat (c) is unknown. The change in temperature (ΔT) is 1.80 °C.
Substitute these values into the formula.
[tex]47.1 \ J = (14.0 \ g) * c * (1.80 \textdegree C)[/tex]
Multiply the 2 numbers in parentheses on the right side of the equation.
[tex]47.1 \ J = (14.0 \ g * 1.80 \textdegree C)*c[/tex]
[tex]47.1 \ J = (25.2 \ g*\textdegree C) *c[/tex]
We are solving for the heat capacity of the liquid, so we must isolate the variable c. It is being multiplied by 25.2 grams * degrees Celsius. The inverse operation of multiplication is division, so we divide both sides of the equation by (25.2 g * °C).
[tex]\frac {47.1 \ J}{(25.2 g *\textdegree C)} = \frac {(25.2 g *\textdegree C)*c}{{(25.2 g *\textdegree C)}}[/tex]
[tex]\frac {47.1 \ J}{(25.2 g *\textdegree C)} =c[/tex]
[tex]1.869047619 \ J/g *\textdegree C = c[/tex]
The original measurements of heat, mass, and temperature all have 3 significant figures, so our answer must have the same. For the number we found that is the hundredth place. The 9 in the thousandth place to the right tells us to round the 6 up to a 7.
[tex]1.87 \ J/ g * \textdegree C =c[/tex]
The heat capacity of the liquid is approximately 1.87 J/g°C.
