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MrRissso [65]
3 years ago
11

How much heat is given off when 869 g iron bar cools from 94 degrees celsius to 5

Chemistry
1 answer:
melomori [17]3 years ago
4 0

Answer:

Heat given off was -34.34kJ

Explanation:

Mass of iron bar = 869g

Initial temperature (T1) = 94°C

Final temperature (T2) = 5°C

Specific heat capacity of iron (c) = 0.444J/g°C

Heat energy (Q) = Mc∇T

Q = heat energy

c = specific heat capacity

∇T = change in temperature

M = mass of the substance

Q = mc∇T

∇T = T2 - T1

Q = Mc(T2 -T1)

Q = 869 * 0.444 * (5 - 94)

Q = 385.836 * -89

Q = -34339.404J

Q = -34.34kJ

The heat given of was -34.34kJ

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3 0
4 years ago
A 825 g iron block is heated to 352 degrees C and is placed in an insulated container (of negligible heat capacity) containing 4
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Answer : The final equilibrium temperature of the water and iron is, 537.12 K

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

q_1=-q_2

m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

where,

c_1 = specific heat of iron =  560 J/(kg.K)

c_1 = specific heat of water = 4186 J/(kg.K)

m_1 = mass of iron = 825 g

m_2 = mass of water = 40 g

T_f = final temperature of water and iron = ?

T_1 = initial temperature of iron = 352^oC=273+352=625K

T_2 = initial temperature of water = 20^oC=273+20=293K

Now put all the given values in the above formula, we get:

(825\times 10^{-3}kg)\times 560J/(kg.K)\times (T_f-625K)=-(40\times 10^{-3}kg)\times 4186J/(kg.K)\times (T_f-293K)

T_f=537.12K

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