Question

12500 J/ (106 g) (4C) reduce to one

Answer 1

The given expression is $\frac{12500 J}{(106g)(4^{0}C )}$

This expression denotes the specific heat of a substance which indicates a heat energy of 12500 J is involved in raising the temperature of 106 g of the substance by $4^{0}C$. Generally, the units of specific heat are $\frac{J}{g.^{0}C }$

$\frac{12500 J}{(106g)(4^{0}C )}$ = $\frac{12500 J}{(106g)(4^{0}C } = 29.5 \frac{J}{g^{0}C }$

Therefore,  $\frac{12500 J}{(106g)(4^{0}C )}$ when reduced to one unit will be 29.5 $\frac{J}{g^{0}C }$

Answer 2

Answer:

29.48 J/g°C

Explanation:

Step 1:

Data obtained from the question. This includes:

12500 J/ (106g) (4°C)

From the question given, we were asked to reduce the expression above to a single value

Step 2:

Reduction of the expression.

The reduction of 12500 J/ (106g) (4°C) can be obtained as follow:

12500 J/ (106g) (4°C) = 12500J/424g°C

= 29.48 J/g°C

Therefore, the reduction of 12500 J/ (106g) (4°C) to a single value is 29.48 J/g°C