The specific heat of ice is 0.492 cal/(g × °C). How many calories of heat are required to raise 100.0 g of ice from -20.0 °C to -0.5 °C?

Jul 7, 2018

Well, your ice DOES NOT undergo a phase change....

Explanation:

And so we simply take the product...

${\text{mass"xx"C"_"specific heat"xxDeltaT=100.0*gxxunderbrace(0.492*cal*g^-1*""^@C^-1)_"quoted specific heat"xx19.5*}}^{\circ} C = + 959.4 \cdot c a l$....the positive sign indicates heat is ADDED to the system.

Jul 7, 2018

$959.4$ calories

Explanation:

We use the specific heat equation, which states that,

$q = m c \Delta T$

where:

• $q$ is the heat energy supplied in joules

• $m$ is the mass of the substance in kilograms

• $c$ is the specific heat capacity of the object in joules

• $\Delta T$ is the change in temperature

Here, $\Delta T = - {0.5}^{\circ} \text{C"-(-20^@"C")=19.5^@"C}$.

So, we get:

$q = 100 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{g"*(0.492 \ "cal")/(color(red)cancelcolor(black)"g"color(red)cancelcolor(black)(""^@"C"))*19.5^@"C}}}}$

$= 959.4 \setminus \text{cal}$