Question

The specific heat capacity (c) of a substance is the thermal energy required
to raise one gram of the substance by one degree Celsius. The total amount
of thermal energy (Q) added to a sample of a substance can be calculated
by multiplying the specific heat capacity (c) by the mass of the sample (m)
and the temperature change (T): Q c m T. It takes 125 kilojoules (kJ)
of energy to heat a cup of water, increasing its temperature by 1°C. What
is the temperature change in °C of the cup of water after 750 kJ of energy
are used? S

Answer 1

Temperature of the same cup of water will rise by 6 °C unless it boils.

Explanation

$Q = c\cdot m \cdot \Delta T$.

However, neither $c$ nor $m$ is given.

Adding $Q = 125\;\text{kJ}$ to this cup of water of mass $m$ rises its temperature by $\Delta T = 1 \; \textdegree{}\text{C}$.

In other words,

$\begin{array}{lll}Q &=& c \cdot m \cdot \Delta T\\125\;\text{kJ}  &=& c \cdot m \times (1 \; \textdegree{}\text{C})\end{array}$

$c \cdot m = \dfrac{Q}{\Delta T} = \dfrac{125\; \text{kJ}}{1 \; \textdegree{}\text{C}}$.

Both $c$ and $m$ are constant for the same cup of water unless the water boils. It's possible to reuse the value of $c \cdot m$ in the second calculation. Here's how:

$\Delta T = \dfrac{Q}{c\cdot m} = \dfrac{750 \; \text{kJ}}{\dfrac{125\;\text{kJ}}{1\;\textdegree{}\text{C}}} = \dfrac{750}{125} \; \textdegree{}\text{C}} = 6 \; \textdegree{}\text{C}}$.