Question

50.0 g of 27.0°C iron is heated until its temperature is 117°C.

How many joules (J) of heat energy (Q) is needed to cause this rise in temperature? _
How many calories (cal) of heat energy (Q) is needed to cause this rise in temperature? _

Answer 1

2025Joules

482.14 calories

Explanation:

Specific heat capacity of iron is 0.450J/g°C

from Q= mass of solid × SHC of solid× change in temperature

Q= 50g × 0.450J/g°C × ( 117°C - 27.0° C)

Q =2025Joules.

from 1 calorie= 4.2 joules

4.2 joules= 1 calorie

2025 joules= (1×2025)/4.2

Q = 482.14 calories