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.

Step-by-step 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 * (1 \; \textdegree{}\text{C})\end{array}`

`c \cdot m = (Q)/(\Delta T) = \frac{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 = (Q)/(c\cdot m) = \frac{750 \; \text{kJ}}{\frac{125\;\text{kJ}}{1\;\textdegree{}\text{C}}} = (750)/(125) \; \textdegree{}\text{C}} = 6 \; \textdegree{}\text{C}}`.