Question

A pan containing 30 grams of water was allowed to cool from a temperature of 90.0 °C. If the amount of heat released is 1,500 joules, what is the approximate final temperature of the water?

A. 76 °C
B. 78 °C
C. 81 °C
D. 82 °C

Answer 1

78 °C

Step-by-step explanation:

I took the test and got it correct. hope this helps

Answer 2

B. 78 °C.

Step-by-step explanation:

• To solve this problem, we can use the relation:

Q = m.c.ΔT,

where, Q is the amount of heat released from water (Q = - 1500 J).

m is the mass of water (m = 30.0 g).

c is the specific heat capacity of water (c = 4.18 J/g.°C).

ΔT is the difference between the initial and final temperature (ΔT = final T - initial T = final T - 90.0 °C).

∴ (- 1500 J) = (30.0 g)(4.18 J/g.°C)(final T - 90.0 °C)

∴ (final T - 90.0 °C) = (- 1500 J)/(30.0 g)(4.18 J/g.°C) = - 11.96°C.

∴ final T = 90.0 °C - 11.96°C = 78.04°C ≅ 78 °C.

So, the right choice is: B. 78 °C.