Question

1kg slab of concrete loses 12,000j of heat when it cools from 30 Celsius to 26 Celsius. Determine the specific heat capacity of concrete.

Answer 1

The specific heat capacity of concrete is $3.0 J/(g^{\circ}C)$

Explanation:

When a certain amount of energy Q is supplied/given off to/from a sample of substance with mass m, the temperature of the substance increases/decreases by an amount $\Delta T$, according to the equation

$Q=mC_s \Delta T$

where

m is the mass of the substance

$C_s$ is the specific heat capacity of the substance

$\Delta T$ is the change in temperature of the substance

In this problem, we have:

m = 1 kg = 1000 g is the mass of the concrete slab

$Q = -12,000 J$ is the amount of energy lost by the slab

$\Delta T = 30-26= -4^{\circ}C$ is the change in temperature of the slab

Solving the equation for $C_s$, we find the specific heat capacity of concrete:

$C_s = \frac{Q}{m \Delta T}=\frac{-12,000}{(1000)(-4)}=3.0 J/(g^{\circ}C)$

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