Question

show work with proper units and sig digs when you answer the following: how much heat in calories is lost if 428 g of water cools from 18.6°c to 17.3°c?

Explanation:

Step1: Define heat loss formula

The formula for heat change is $q = mc\Delta T$, where $m$ = mass, $c$ = specific heat of water ($1\ \text{cal/g}^\circ\text{C}$), $\Delta T = T_{\text{final}} - T_{\text{initial}}$.

Step2: Calculate temperature change

$\Delta T = 17.3^\circ\text{C} - 18.6^\circ\text{C} = -1.3^\circ\text{C}$

Step3: Plug values into formula

$q = 428\ \text{g} \times 1\ \frac{\text{cal}}{\text{g}^\circ\text{C}} \times (-1.3^\circ\text{C})$

Step4: Compute heat loss

$q = -556.4\ \text{cal}$ (negative sign indicates heat loss)

Step5: Round to sig figs

Round to 2 significant figures (from $\Delta T$): $q = -560\ \text{cal}$

Answer:

$560$ calories of heat are lost (or $-560\ \text{cal}$)