Question

A 250-g block of ice is removed from the refrigerator at -8.0°C. How much thermal energy does the ice absorb as it warms to room temperature (22°C)? The heat of fusion of water is 3.34x10^5 J/kg. )(C ice=2060 J/(kg • c))( C water 4180 J/(kg • c)

Answer 1

Give;:

Mass of block = 250 g

Tempearture of refrigerator = -8.0°C

heat of fusion of water is 3.34x10^5 J/kg. c

specific heat of water = 4180 J/kg

Let's find the thermal energy the ice absorbs as it warns th room temperature of 22°C).

Apply the formula:

`Q=mc(T_2-T_1)+h_fm_{}`

Where:

c = 4180 J/kg

m = 250 g = 0.2450kg

T2 = 22°C

T1 = -8.0°C

hf = 3.34x10⁵ J/kg. c

Thus, we have:

`\begin{gathered} Q=0.250\ast4180(22-(-8.0))+(3.34\ast10^5\ast0.250) \\ \\ Q=0.250\ast4180(22+8.0))+(3.34\ast10^5\ast0.250) \\ \\ Q=1045(30.0)+(83500) \\ \\ Q=114850\text{ J} \end{gathered}`

Therefore, the thermal energy the ice absorbs is 114850 Joules.

ANSWER:

114850 J