When 168 joules of heat is added 4 grams of water at 283 K, what is the resulting temperature?

May 11, 2017

$293 K$

Explanation:

The specific heat formula:
$Q = c \cdot m \cdot \Delta T$, where $Q$ is the amount of heat transferred, $c$ is the specific heat capacity of the substance, $m$ is the mass of the object, and $\Delta T$ is the change in temperature. In order to solve for the change in temperature, use the formula

$\Delta T = \frac{Q}{{c}_{w a t e r} \cdot m}$

The standard heat capacity of water, ${c}_{w a t e r}$ is $4.18 \cdot J \cdot {g}^{- 1} \cdot {K}^{- 1}$.

And we get $\Delta T = \frac{168 \cdot J}{4.18 \cdot J \cdot {g}^{- 1} \cdot {K}^{- 1} \cdot 4 \cdot g} = 10.0 K$

Since $Q > 0$, the resulting temperature will be ${T}_{f} = {T}_{i} + \Delta T = 283 K + 10.0 K = 293 K$
(pay special attention to significant figures)

Additional resources on Heat Capacity and Specific Heat:
https://www.ck12.org/chemistry/Heat-Capacity-and-Specific-Heat/lesson/Heat-Capacity-and-Specific-Heat-CHEM/?referrer=concept_details