Question

a 2.40 kg block of ice is heated with 5820 j of heat. the specific heat of ice is 2.10 j·g⁻¹·°c⁻¹. by how much will its temperature rise, assuming it does not melt?
a 15,500,000 °c
b 15,500 °c
c 115 °c
d 1.15 °c

Explanation:

Step1: Convert mass to grams

Given mass \(m = 2.40\space kg=2.40\times10^{3}\space g\) (since \(1\space kg = 1000\space g\)).

Step2: Use the heat - transfer formula \(Q = mc\Delta T\)

We need to solve for \(\Delta T\). Rearranging the formula \(Q = mc\Delta T\) gives \(\Delta T=\frac{Q}{mc}\).
Substitute \(Q = 5820\space J\), \(m = 2.40\times 10^{3}\space g\), and \(c=2.10\space J\cdot g^{-1}\cdot^{\circ}C^{-1}\) into the formula:

$$
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$$

Answer:

D. \(1.15^{\circ}C\)